💡Alright, so I have this problem here about a function ( f(x) = ln x - frac{m(x + n)}{x + 1} ) where ( m > 0 ) and ( n ) is a real number. The problem states that this function is not monotonically increasing on the interval ( (0, +infty) ). Then, it asks for the range of the real number ( lambda ) such that the inequality ( m - n > lambda ) always holds. The options are given as A through D.First, I need to understand what it means for a function to be monotonically increasing. A function is monotonically increasing if its derivative is non-negative throughout the interval. Since the function is not monotonically increasing, that means its derivative must be negative somewhere in the interval ( (0, +infty) ). So, I should probably start by finding the derivative of ( f(x) ).Let me compute the derivative ( f'(x) ). The function ( f(x) ) has two parts: ( ln x ) and ( -frac{m(x + n)}{x + 1} ). The derivative of ( ln x ) is ( frac{1}{x} ). For the second part, I'll use the quotient rule. The derivative of ( frac{m(x + n)}{x + 1} ) is:Let me denote ( u = m(x + n) ) and ( v = x + 1 ). Then, ( u' = m ) and ( v' = 1 ). The quotient rule says that ( frac{d}{dx} frac{u}{v} = frac{u'v - uv'}{v^2} ). Plugging in, we get:[frac{d}{dx} left( frac{m(x + n)}{x + 1} right) = frac{m(x + 1) - m(x + n)}{(x + 1)^2} = frac{m(x + 1 - x - n)}{(x + 1)^2} = frac{m(1 - n)}{(x + 1)^2}]So, the derivative of the entire function ( f(x) ) is:[f'(x) = frac{1}{x} - frac{m(1 - n)}{(x + 1)^2}]Wait, that doesn't seem right. Let me double-check. The derivative of ( f(x) ) is the derivative of ( ln x ) minus the derivative of ( frac{m(x + n)}{x + 1} ). So, it should be:[f'(x) = frac{1}{x} - frac{m(1 - n)}{(x + 1)^2}]Yes, that looks correct.Now, since the function is not monotonically increasing, there must be some ( x ) in ( (0, +infty) ) where ( f'(x) < 0 ). So, I need to find when:[frac{1}{x} - frac{m(1 - n)}{(x + 1)^2} < 0]Let me rearrange this inequality:[frac{1}{x} < frac{m(1 - n)}{(x + 1)^2}]Multiply both sides by ( x(x + 1)^2 ) (which is positive since ( x > 0 )):[(x + 1)^2 < m(1 - n)x]So, we have:[(x + 1)^2 - m(1 - n)x < 0]Let me expand ( (x + 1)^2 ):[x^2 + 2x + 1 - m(1 - n)x < 0]Combine like terms:[x^2 + (2 - m(1 - n))x + 1 < 0]So, this is a quadratic in ( x ):[x^2 + [2 - m(1 - n)]x + 1 < 0]For this quadratic to be negative for some ( x > 0 ), it must have real roots, and the quadratic must open upwards (since the coefficient of ( x^2 ) is positive). Therefore, the quadratic must have two distinct real roots, and the interval between them must lie within ( (0, +infty) ).The condition for a quadratic ( ax^2 + bx + c ) to have two distinct real roots is that its discriminant ( D ) must be positive:[D = b^2 - 4ac > 0]In our case, ( a = 1 ), ( b = 2 - m(1 - n) ), and ( c = 1 ). So,[D = [2 - m(1 - n)]^2 - 4(1)(1) > 0]Simplify:[[2 - m(1 - n)]^2 - 4 > 0]Let me expand ( [2 - m(1 - n)]^2 ):[4 - 4m(1 - n) + m^2(1 - n)^2 - 4 > 0]Simplify:[-4m(1 - n) + m^2(1 - n)^2 > 0]Factor out ( m(1 - n) ):[m(1 - n)[m(1 - n) - 4] > 0]So, we have:[m(1 - n)[m(1 - n) - 4] > 0]Since ( m > 0 ), the sign of the expression depends on ( (1 - n) ) and ( [m(1 - n) - 4] ).Let me denote ( k = m(1 - n) ). Then, the inequality becomes:[k(k - 4) > 0]This is a quadratic inequality. The product is positive when both factors are positive or both are negative.Case 1: ( k > 0 ) and ( k - 4 > 0 ) ⇒ ( k > 4 )Case 2: ( k < 0 ) and ( k - 4 < 0 ) ⇒ ( k < 0 )But ( k = m(1 - n) ). Since ( m > 0 ), ( k > 0 ) implies ( 1 - n > 0 ) ⇒ ( n < 1 ). Similarly, ( k < 0 ) implies ( 1 - n < 0 ) ⇒ ( n > 1 ).So, we have two cases:1. ( n < 1 ) and ( k > 4 ) ⇒ ( m(1 - n) > 4 )2. ( n > 1 ) and ( k < 0 ) ⇒ ( m(1 - n) < 0 )But ( m > 0 ), so in the second case, ( 1 - n < 0 ) ⇒ ( n > 1 ), which is consistent.Now, let's analyze both cases.Case 1: ( n < 1 ) and ( m(1 - n) > 4 )We need to find the range of ( lambda ) such that ( m - n > lambda ) always holds.Let me express ( m - n ) in terms of ( k ). Since ( k = m(1 - n) ), we can write ( m = frac{k}{1 - n} ). Therefore,[m - n = frac{k}{1 - n} - n = frac{k - n(1 - n)}{1 - n}]But this seems complicated. Maybe another approach.Let me consider ( m - n ). From ( k = m(1 - n) ), we have ( m = frac{k}{1 - n} ). So,[m - n = frac{k}{1 - n} - n = frac{k - n(1 - n)}{1 - n} = frac{k - n + n^2}{1 - n}]Hmm, not sure if that helps. Maybe instead, express ( m - n ) in terms of ( k ):Since ( k = m(1 - n) ), then ( m = frac{k}{1 - n} ). So,[m - n = frac{k}{1 - n} - n = frac{k - n(1 - n)}{1 - n} = frac{k - n + n^2}{1 - n}]Alternatively, maybe express ( m - n ) as ( m - n = m - n ). Since ( k = m(1 - n) ), perhaps express ( m - n ) in terms of ( k ).Wait, let's think differently. Since ( m(1 - n) > 4 ) in Case 1, and ( n < 1 ), let's denote ( t = 1 - n ), so ( t > 0 ). Then, ( m t > 4 ), so ( m > frac{4}{t} ).We need to find the minimum value of ( m - n ). Since ( n = 1 - t ), ( m - n = m - (1 - t) = m + t - 1 ).Given ( m > frac{4}{t} ), so ( m + t - 1 > frac{4}{t} + t - 1 ).Let me denote ( g(t) = frac{4}{t} + t - 1 ) for ( t > 0 ). To find the minimum of ( g(t) ), take derivative:[g'(t) = -frac{4}{t^2} + 1]Set ( g'(t) = 0 ):[-frac{4}{t^2} + 1 = 0 ⇒ frac{4}{t^2} = 1 ⇒ t^2 = 4 ⇒ t = 2 ) (since ( t > 0 ))So, the minimum occurs at ( t = 2 ). Compute ( g(2) = frac{4}{2} + 2 - 1 = 2 + 2 - 1 = 3 ).Therefore, ( m - n > 3 ) in Case 1.Case 2: ( n > 1 ) and ( m(1 - n) < 0 ). Since ( m > 0 ), this implies ( 1 - n < 0 ) ⇒ ( n > 1 ), which is consistent.But in this case, ( m(1 - n) < 0 ). However, ( m - n ) is ( m - n ). Since ( n > 1 ), ( m - n ) could be positive or negative depending on ( m ). But we need ( m - n > lambda ) always holds. However, in this case, ( m ) is positive, but ( n ) is greater than 1, so ( m - n ) could be less than any real number, but we need ( m - n > lambda ) always. This seems conflicting because if ( n ) can be very large, ( m - n ) can be very negative, which would violate ( m - n > lambda ) for any finite ( lambda ). Therefore, Case 2 might not contribute to the range of ( lambda ) because we cannot have ( m - n > lambda ) always hold if ( n ) can be arbitrarily large.Therefore, the only valid case is Case 1, where ( m - n > 3 ). Hence, ( lambda ) must be less than 3 for ( m - n > lambda ) to always hold.Wait, but the options are:A. [3, +∞)B. [4, +∞)C. (-∞, 3)D. (-∞, 4)So, if ( m - n > 3 ), then ( lambda ) must be less than 3 to satisfy ( m - n > lambda ) always. Therefore, the range of ( lambda ) is ( (-infty, 3) ), which is option C.But wait, let me double-check. If ( m - n > 3 ), then ( lambda ) must be less than or equal to 3 to ensure ( m - n > lambda ). But the problem says "the inequality ( m - n > lambda ) always holds". So, ( lambda ) must be less than the minimum value of ( m - n ), which is 3. Therefore, ( lambda ) must be less than 3, so the range is ( (-infty, 3) ).Yes, that makes sense. So, the correct answer should be C.

💡Okay, so I have this problem about a regular hexagon with side length 3, and I need to find the area of triangle ACE. Hmm, let me think about how to approach this.First, I remember that a regular hexagon can be divided into six equilateral triangles, all with side length equal to the side length of the hexagon. So, in this case, each of those triangles has sides of length 3. That might be useful.Now, the hexagon is labeled ABCDEF, so the vertices are A, B, C, D, E, F in order. Triangle ACE is formed by connecting every other vertex, right? So, A to C, C to E, and E back to A. That should form a triangle inside the hexagon.I think it might help to visualize or maybe even sketch the hexagon. Let me imagine it: each internal angle of a regular hexagon is 120 degrees, and each side is equal. So, connecting A to C skips one vertex, which is B, and similarly, C to E skips D, and E back to A skips F. So, triangle ACE is a larger triangle inside the hexagon.I wonder if triangle ACE is equilateral. Since all sides of the hexagon are equal, maybe the sides of triangle ACE are equal too. Let me check that. The distance from A to C, C to E, and E to A should be the same.In a regular hexagon, the distance between two non-adjacent vertices can be calculated. For example, the distance from A to C is the length of the diagonal that skips one vertex. I think in a regular hexagon, the length of such a diagonal is twice the side length times the sine of 60 degrees, because each internal angle is 120 degrees, and the central angles are 60 degrees.Wait, maybe I should recall the formula for the length of a diagonal in a regular polygon. The formula is ( d = 2 times text{side} times sin(pi/n) ), where ( n ) is the number of sides. For a hexagon, ( n = 6 ), so ( d = 2 times 3 times sin(pi/6) ). But ( sin(pi/6) ) is 0.5, so ( d = 2 times 3 times 0.5 = 3 ). Wait, that can't be right because the side length is 3, and the diagonal should be longer than the side length.Hmm, maybe I used the wrong formula. Let me think again. In a regular hexagon, the distance between two vertices separated by one other vertex (like A to C) is actually equal to the side length multiplied by ( sqrt{3} ). Because in the equilateral triangles that make up the hexagon, the height is ( frac{sqrt{3}}{2} times text{side length} ). So, maybe the diagonal AC is ( 3 times sqrt{3} ).Wait, no, that's the height of the equilateral triangle. Maybe I should use the law of cosines to find the length of AC. In triangle ABC, which is an equilateral triangle, angle at B is 60 degrees, but in the hexagon, angle at B is 120 degrees. Hmm, maybe I need to consider triangle ABC in the hexagon.Wait, no, triangle ABC is actually part of the hexagon, but it's not an equilateral triangle. Wait, no, in a regular hexagon, each internal angle is 120 degrees, so triangle ABC is actually an isosceles triangle with two sides equal to 3 and the included angle of 120 degrees. So, using the law of cosines, the length of AC can be calculated.Yes, that makes more sense. So, in triangle ABC, sides AB and BC are both 3, and the angle at B is 120 degrees. So, the length of AC is:( AC^2 = AB^2 + BC^2 - 2 times AB times BC times cos(120^circ) )Plugging in the values:( AC^2 = 3^2 + 3^2 - 2 times 3 times 3 times cos(120^circ) )( AC^2 = 9 + 9 - 18 times cos(120^circ) )I know that ( cos(120^circ) = -frac{1}{2} ), so:( AC^2 = 18 - 18 times (-frac{1}{2}) )( AC^2 = 18 + 9 = 27 )So, ( AC = sqrt{27} = 3sqrt{3} )Okay, so the length of AC is ( 3sqrt{3} ). Similarly, the lengths of CE and EA should also be ( 3sqrt{3} ) because the hexagon is regular, so all such diagonals are equal.So, triangle ACE is an equilateral triangle with sides of length ( 3sqrt{3} ). Therefore, the area of an equilateral triangle is given by the formula:( text{Area} = frac{sqrt{3}}{4} times text{side}^2 )Plugging in the side length:( text{Area} = frac{sqrt{3}}{4} times (3sqrt{3})^2 )Calculating ( (3sqrt{3})^2 ):( (3sqrt{3})^2 = 9 times 3 = 27 )So,( text{Area} = frac{sqrt{3}}{4} times 27 = frac{27sqrt{3}}{4} )Wait, but I'm not sure if triangle ACE is equilateral. Let me double-check. In the regular hexagon, the central angles are 60 degrees, so the angle at the center between two adjacent vertices is 60 degrees. But in triangle ACE, the angle at each vertex might not be 60 degrees.Wait, actually, in the regular hexagon, the angle at each vertex is 120 degrees, but when you connect every other vertex, the internal angles of triangle ACE might be different. Maybe I made a mistake assuming it's equilateral.Let me think again. If I consider the regular hexagon inscribed in a circle, all the vertices lie on the circumference. The central angles between each vertex are 60 degrees because 360/6=60. So, the angle subtended by each side at the center is 60 degrees.Now, triangle ACE connects every other vertex, so the central angles between A and C is 120 degrees, between C and E is another 120 degrees, and between E and A is another 120 degrees. Wait, no, that can't be because 120*3=360, which is the full circle.Wait, actually, each side of the hexagon subtends 60 degrees at the center, so the angle between A and C would be 120 degrees because they are two sides apart. Similarly, between C and E is another 120 degrees, and between E and A is another 120 degrees. So, the central angles are all 120 degrees.But triangle ACE is formed by connecting these points, so the triangle is actually equilateral because all sides are equal (each being 2 radii times sin(60), but wait, no, earlier I calculated AC as 3√3, which would be the length if the radius is 3. Wait, maybe I need to clarify.Alternatively, maybe I should consider the coordinates of the hexagon vertices and calculate the area using coordinates.Let me place the regular hexagon on a coordinate system with the center at the origin. Let's assume the hexagon is regular and centered at (0,0). Let me assign coordinates to each vertex.Starting with vertex A at (3,0). Then, moving counterclockwise, each vertex will be separated by 60 degrees. So, the coordinates can be calculated using cosine and sine for each angle.So, vertex A: (3,0)Vertex B: (3 cos 60°, 3 sin 60°) = (1.5, (3√3)/2 )Vertex C: (3 cos 120°, 3 sin 120°) = (-1.5, (3√3)/2 )Vertex D: (-3,0)Vertex E: (3 cos 240°, 3 sin 240°) = (-1.5, -(3√3)/2 )Vertex F: (3 cos 300°, 3 sin 300°) = (1.5, -(3√3)/2 )So, now, triangle ACE has vertices at A(3,0), C(-1.5, (3√3)/2 ), and E(-1.5, -(3√3)/2 ).Now, to find the area of triangle ACE, I can use the shoelace formula.The shoelace formula is:Area = |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))/2|Plugging in the coordinates:x1 = 3, y1 = 0x2 = -1.5, y2 = (3√3)/2x3 = -1.5, y3 = -(3√3)/2So,Area = |(3*((3√3)/2 - (-(3√3)/2)) + (-1.5)*(-(3√3)/2 - 0) + (-1.5)*(0 - (3√3)/2))/2|Let me compute each term step by step.First term: 3*((3√3)/2 - (-(3√3)/2)) = 3*((3√3)/2 + (3√3)/2) = 3*(6√3/2) = 3*(3√3) = 9√3Second term: (-1.5)*(-(3√3)/2 - 0) = (-1.5)*(-3√3/2) = (1.5)*(3√3/2) = (3/2)*(3√3/2) = (9√3)/4Third term: (-1.5)*(0 - (3√3)/2) = (-1.5)*(-3√3/2) = (1.5)*(3√3/2) = (3/2)*(3√3/2) = (9√3)/4Now, adding all three terms:9√3 + (9√3)/4 + (9√3)/4Convert 9√3 to quarters: 9√3 = 36√3/4So, total sum: 36√3/4 + 9√3/4 + 9√3/4 = (36 + 9 + 9)√3/4 = 54√3/4Now, take the absolute value (which is positive) and divide by 2:Area = |54√3/4| / 2 = (54√3/4) / 2 = 54√3/8 = 27√3/4Wait, that's the same result as before, 27√3/4. But earlier I thought maybe I made a mistake assuming it's equilateral. But according to the shoelace formula, it's 27√3/4.But wait, earlier when I calculated AC as 3√3, and then thought of it as an equilateral triangle, I got 27√3/4 as well. So, maybe it is equilateral after all.Wait, but in the coordinate system, the sides AC, CE, and EA are all equal to 3√3, so triangle ACE is indeed equilateral. Therefore, the area is 27√3/4.But wait, the problem says to express the answer in simplest radical form, so 27√3/4 is already simplified.Wait, but earlier I thought maybe I made a mistake because I was confused about the angles, but both methods gave me the same answer, so I think it's correct.Alternatively, another way to think about it is that the regular hexagon can be divided into six equilateral triangles, each with side length 3. The area of the hexagon is 6 times the area of one such triangle.Area of one equilateral triangle: (√3/4)*3² = (√3/4)*9 = 9√3/4So, total area of hexagon: 6*(9√3/4) = 54√3/4 = 27√3/2Now, triangle ACE is a larger triangle inside the hexagon. How much of the hexagon does it cover? Well, if I look at the hexagon divided into six equilateral triangles, triangle ACE would cover every other triangle, so maybe three of them? But wait, no, because triangle ACE is formed by connecting every other vertex, it actually covers more area.Wait, actually, when you connect A, C, E in the hexagon, you're forming a triangle that encompasses the center and three of the six equilateral triangles. So, the area of triangle ACE would be equal to the area of three of those small equilateral triangles.But wait, each small triangle has area 9√3/4, so three of them would be 27√3/4, which matches the earlier result.So, that confirms it again. Therefore, the area of triangle ACE is 27√3/4.Wait, but I'm a bit confused because earlier I thought triangle ACE was equilateral, but in the coordinate system, the triangle has vertices at (3,0), (-1.5, (3√3)/2), and (-1.5, -(3√3)/2). Plotting these points, it seems like the triangle is symmetric about the x-axis, with a base from (-1.5, (3√3)/2) to (-1.5, -(3√3)/2), which is a vertical line at x=-1.5, and the third vertex at (3,0). So, actually, the triangle is not equilateral but isosceles.Wait, that contradicts my earlier conclusion. Hmm, maybe I made a mistake in assuming it's equilateral.Wait, let me recast the coordinates:A is at (3,0)C is at (-1.5, (3√3)/2 )E is at (-1.5, -(3√3)/2 )So, the sides:AC: distance from (3,0) to (-1.5, (3√3)/2 )Using distance formula:√[(3 - (-1.5))² + (0 - (3√3)/2 )²] = √[(4.5)² + ( (3√3)/2 )²] = √[20.25 + (27/4)] = √[20.25 + 6.75] = √[27] = 3√3Similarly, CE: distance from (-1.5, (3√3)/2 ) to (-1.5, -(3√3)/2 )That's just the vertical distance: (3√3)/2 - (-(3√3)/2 ) = 3√3And EA: distance from (-1.5, -(3√3)/2 ) to (3,0)Same as AC: √[(3 - (-1.5))² + (0 - (-(3√3)/2 ))²] = same as AC, which is 3√3So, all sides are equal, so triangle ACE is indeed equilateral with side length 3√3.Therefore, the area is (√3/4)*(3√3)^2 = (√3/4)*(27) = 27√3/4.So, despite the initial confusion about the shape, the calculations confirm that it's equilateral and the area is 27√3/4.Alternatively, another approach is to note that triangle ACE divides the hexagon into four smaller congruent equilateral triangles. Wait, no, actually, when you draw triangle ACE, it divides the hexagon into three smaller congruent rhombuses and the central triangle ACE.But regardless, the area calculation via coordinates and via the equilateral triangle formula both give the same result, so I'm confident that 27√3/4 is the correct answer.Wait, but let me double-check the shoelace formula calculation to make sure I didn't make any arithmetic errors.Using the coordinates:A(3,0), C(-1.5, (3√3)/2 ), E(-1.5, -(3√3)/2 )Shoelace formula:Area = |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))/2|Plugging in:x1 = 3, y1 = 0x2 = -1.5, y2 = (3√3)/2x3 = -1.5, y3 = -(3√3)/2Compute each term:First term: x1(y2 - y3) = 3*((3√3)/2 - (-(3√3)/2)) = 3*( (3√3)/2 + (3√3)/2 ) = 3*(6√3/2) = 3*3√3 = 9√3Second term: x2(y3 - y1) = (-1.5)*( -(3√3)/2 - 0 ) = (-1.5)*(-3√3/2) = (1.5)*(3√3/2) = (3/2)*(3√3/2) = 9√3/4Third term: x3(y1 - y2) = (-1.5)*(0 - (3√3)/2 ) = (-1.5)*(-3√3/2) = (1.5)*(3√3/2) = same as second term, 9√3/4Adding them up: 9√3 + 9√3/4 + 9√3/4Convert 9√3 to quarters: 9√3 = 36√3/4So total: 36√3/4 + 9√3/4 + 9√3/4 = (36 + 9 + 9)√3/4 = 54√3/4Divide by 2: 54√3/4 / 2 = 54√3/8 = 27√3/4Yes, that's correct. So, the area is indeed 27√3/4.Another way to think about it is that the regular hexagon can be divided into six equilateral triangles, each with side length 3. The area of each such triangle is (√3/4)*3² = 9√3/4. The total area of the hexagon is 6*(9√3/4) = 54√3/4 = 27√3/2.Now, triangle ACE is a larger triangle inside the hexagon. How much of the hexagon does it cover? If I consider the hexagon divided into six equilateral triangles, triangle ACE would cover three of them, but actually, it's more than that because it's a larger triangle encompassing the center.Wait, no, when you connect A, C, E, you're essentially forming a triangle that covers the center and three of the six small triangles. But actually, each side of triangle ACE cuts through two small triangles, so it might cover more.Alternatively, perhaps triangle ACE is made up of three of the small equilateral triangles and three rhombuses. But regardless, the area calculation via coordinates and via the equilateral triangle formula both give 27√3/4, which is exactly half of the hexagon's area (since the hexagon's area is 27√3/2). That makes sense because triangle ACE divides the hexagon into two equal areas: the triangle itself and the three surrounding rhombuses which together make up the other half.So, yes, the area of triangle ACE is half the area of the hexagon, which is 27√3/4.Therefore, after multiple methods and checks, I'm confident that the area of triangle ACE is 27√3/4.

💡Okay, so I have this problem here: I need to find the minimum value of the expression ( x^2 + y^2 - 8x + 6y + 20 ) for real numbers ( x ) and ( y ). Hmm, I remember that when dealing with quadratic expressions, completing the square can be a useful technique to find minima or maxima. Let me try to recall how that works.First, I think I should group the terms involving ( x ) together and the terms involving ( y ) together. So, let me rewrite the expression:( x^2 - 8x + y^2 + 6y + 20 )Alright, now I have the ( x ) terms and the ( y ) terms separated. I think the next step is to complete the square for both ( x ) and ( y ) separately. Let me start with the ( x ) terms: ( x^2 - 8x ).To complete the square, I remember that I take the coefficient of ( x ), which is -8, divide it by 2 to get -4, and then square it to get 16. So, I can write:( x^2 - 8x = (x - 4)^2 - 16 )Wait, let me check that. If I expand ( (x - 4)^2 ), I get ( x^2 - 8x + 16 ). So, to make it equal to ( x^2 - 8x ), I need to subtract 16. Yeah, that makes sense.Now, moving on to the ( y ) terms: ( y^2 + 6y ). Similarly, I take the coefficient of ( y ), which is 6, divide by 2 to get 3, and square it to get 9. So, I can write:( y^2 + 6y = (y + 3)^2 - 9 )Let me verify that. Expanding ( (y + 3)^2 ) gives ( y^2 + 6y + 9 ). So, subtracting 9 gives me back ( y^2 + 6y ). Perfect.Now, substituting these back into the original expression:( (x - 4)^2 - 16 + (y + 3)^2 - 9 + 20 )Let me simplify the constants:-16 - 9 + 20 = (-25) + 20 = -5So, the expression becomes:( (x - 4)^2 + (y + 3)^2 - 5 )Hmm, now I have the expression in terms of squares of ( (x - 4) ) and ( (y + 3) ). Since squares are always non-negative, the smallest value each square can take is 0. Therefore, the minimum value of the entire expression occurs when both ( (x - 4)^2 ) and ( (y + 3)^2 ) are zero.So, setting ( (x - 4)^2 = 0 ) gives ( x = 4 ), and setting ( (y + 3)^2 = 0 ) gives ( y = -3 ).Substituting these values back into the expression:( (4 - 4)^2 + (-3 + 3)^2 - 5 = 0 + 0 - 5 = -5 )Therefore, the minimum value of the expression is -5, achieved when ( x = 4 ) and ( y = -3 ).Wait a second, let me make sure I didn't make any mistakes. I completed the square correctly for both ( x ) and ( y ), right? For ( x^2 - 8x ), I added and subtracted 16, and for ( y^2 + 6y ), I added and subtracted 9. Then, combining the constants: -16 -9 +20 is indeed -5. Yeah, that seems right.Also, since both squared terms are non-negative, the smallest they can be is zero, so the minimum value is indeed -5. I think I did everything correctly. Maybe I can double-check by plugging in some other values. Let's say ( x = 4 ) and ( y = -3 ), then the original expression becomes:( 4^2 + (-3)^2 - 8*4 + 6*(-3) + 20 )Calculating each term:16 + 9 - 32 - 18 + 20Adding them up:16 + 9 = 2525 - 32 = -7-7 - 18 = -25-25 + 20 = -5Yep, that checks out. What if I choose another point, say ( x = 5 ) and ( y = -2 )?Plugging into the original expression:25 + 4 - 40 - 12 + 2025 + 4 = 2929 - 40 = -11-11 -12 = -23-23 +20 = -3Which is higher than -5, so that makes sense. How about ( x = 3 ) and ( y = -4 )?9 + 16 -24 -24 +209 +16=2525 -24=11 -24=-23-23 +20=-3Again, higher than -5. So, it seems like -5 is indeed the minimum value.I think I'm confident with this answer now. The minimum value is -5, achieved at ( x = 4 ) and ( y = -3 ).

💡Okay, so I have this problem: the function f(x) = x² + mx - 1 is monotonic on the interval [-1, 3]. I need to find the range of the real number m. Hmm, let me think about how to approach this.First, I remember that a quadratic function like f(x) = ax² + bx + c has a graph that's a parabola. Since the coefficient of x² is 1, which is positive, the parabola opens upwards. That means it has a minimum point, and it's decreasing to the left of the vertex and increasing to the right of the vertex.So, the vertex of this parabola is at x = -b/(2a). In this case, a is 1 and b is m, so the x-coordinate of the vertex is -m/(2*1) = -m/2. That's the axis of symmetry. So, the function is decreasing on (-∞, -m/2) and increasing on (-m/2, ∞).Now, the problem says that f(x) is monotonic on the interval [-1, 3]. Monotonic means the function is either entirely non-increasing or non-decreasing on that interval. Since the parabola opens upwards, it can only be monotonic on an interval if that interval is entirely to the left or entirely to the right of the vertex.So, for f(x) to be monotonic on [-1, 3], the entire interval must lie either to the left of the vertex or to the right of the vertex. That means the vertex must be either to the left of -1 or to the right of 3.Let me write that down:1. If the vertex is to the left of -1, then -m/2 ≤ -1.2. If the vertex is to the right of 3, then -m/2 ≥ 3.Let me solve these inequalities for m.Starting with the first inequality:-m/2 ≤ -1Multiply both sides by 2:-m ≤ -2Multiply both sides by -1, remembering to reverse the inequality:m ≥ 2Now the second inequality:-m/2 ≥ 3Multiply both sides by 2:-m ≥ 6Multiply both sides by -1, reversing the inequality:m ≤ -6So, combining these two results, m must be either greater than or equal to 2 or less than or equal to -6.Therefore, the range of m is (-∞, -6] ∪ [2, ∞).Wait, let me double-check. If m is 2, then the vertex is at -2/2 = -1. So, the vertex is at -1, which is the left endpoint of the interval. Since the function is decreasing to the left of the vertex and increasing to the right, on the interval [-1, 3], the function would start at the vertex and increase. So, it's monotonic increasing on [-1, 3]. That makes sense.Similarly, if m is -6, then the vertex is at -(-6)/2 = 3. So, the vertex is at 3, the right endpoint. The function is decreasing to the left of the vertex, so on the interval [-1, 3], it would be decreasing. That also makes sense.If m is between -6 and 2, say m = 0, then the vertex is at 0. So, the function would decrease from -1 to 0 and then increase from 0 to 3. That means it's not monotonic on the entire interval [-1, 3], which is why m can't be between -6 and 2.So, I think my solution is correct. The range of m is all real numbers less than or equal to -6 and greater than or equal to 2.**Final Answer**The range of the real number ( m ) is boxed{(-infty, -6] cup [2, +infty)}.

💡Okay, so I have this inequality to prove: frac{a^{5}-1}{a^{4}-1} cdot frac{b^{5}-1}{b^{4}-1} > frac{25}{64}(a+1)(b+1)where ( a ) and ( b ) are positive numbers, and neither ( a ) nor ( b ) is equal to 1. Hmm, interesting. I need to show that the product of these two fractions is greater than ( frac{25}{64} ) times the product of ( (a+1) ) and ( (b+1) ). First, I should probably try to simplify each fraction separately. Let me focus on the first fraction: ( frac{a^{5}-1}{a^{4}-1} ). I remember that expressions like ( a^n - 1 ) can often be factored using the formula for the sum of a geometric series. Specifically, ( a^n - 1 = (a - 1)(a^{n-1} + a^{n-2} + dots + a + 1) ). Applying that to both the numerator and the denominator:- The numerator ( a^5 - 1 ) can be factored as ( (a - 1)(a^4 + a^3 + a^2 + a + 1) ).- The denominator ( a^4 - 1 ) can be factored as ( (a - 1)(a^3 + a^2 + a + 1) ).So, when I take the ratio:frac{a^{5}-1}{a^{4}-1} = frac{(a - 1)(a^4 + a^3 + a^2 + a + 1)}{(a - 1)(a^3 + a^2 + a + 1)}Since ( a neq 1 ), I can cancel out the ( (a - 1) ) terms:frac{a^{5}-1}{a^{4}-1} = frac{a^4 + a^3 + a^2 + a + 1}{a^3 + a^2 + a + 1}Okay, so that simplifies the first fraction. Similarly, the second fraction ( frac{b^{5}-1}{b^{4}-1} ) simplifies in the same way:frac{b^{5}-1}{b^{4}-1} = frac{b^4 + b^3 + b^2 + b + 1}{b^3 + b^2 + b + 1}So now, the original inequality becomes:left( frac{a^4 + a^3 + a^2 + a + 1}{a^3 + a^2 + a + 1} right) cdot left( frac{b^4 + b^3 + b^2 + b + 1}{b^3 + b^2 + b + 1} right) > frac{25}{64}(a+1)(b+1)Hmm, so I need to show that the product of these two fractions is greater than ( frac{25}{64}(a+1)(b+1) ). Maybe I can analyze each fraction separately and then combine the results.Let me consider just the first fraction:frac{a^4 + a^3 + a^2 + a + 1}{a^3 + a^2 + a + 1}I wonder if I can find a lower bound for this expression. Maybe I can compare it to some multiple of ( a + 1 ). Let me try to manipulate this expression.Let me denote the numerator as ( N = a^4 + a^3 + a^2 + a + 1 ) and the denominator as ( D = a^3 + a^2 + a + 1 ). So, the fraction is ( frac{N}{D} ).I can write ( N = a cdot D + 1 ), because:( a cdot D = a(a^3 + a^2 + a + 1) = a^4 + a^3 + a^2 + a ), and then adding 1 gives ( N = a^4 + a^3 + a^2 + a + 1 ). So, indeed, ( N = aD + 1 ).Therefore, ( frac{N}{D} = a + frac{1}{D} ).So, ( frac{a^4 + a^3 + a^2 + a + 1}{a^3 + a^2 + a + 1} = a + frac{1}{a^3 + a^2 + a + 1} ).Hmm, that's interesting. So, the fraction is equal to ( a ) plus the reciprocal of ( D ). Since ( a ) is positive and ( D ) is positive, ( frac{1}{D} ) is positive, so ( frac{N}{D} > a ).But that might not be directly helpful. Maybe I can find a better lower bound. Let me try to express ( frac{N}{D} ) in terms of ( a + 1 ).Alternatively, perhaps I can use some inequality like AM-GM or Cauchy-Schwarz to bound this expression.Wait, another idea: Maybe I can write both ( N ) and ( D ) in terms of ( (a + 1) ) and some other terms. Let me see.Alternatively, perhaps I can express ( N ) and ( D ) as polynomials and perform some division or comparison.Wait, another approach: Let me consider the function ( f(a) = frac{a^4 + a^3 + a^2 + a + 1}{a^3 + a^2 + a + 1} ). Maybe I can analyze its behavior for ( a > 0 ), ( a neq 1 ).Let me compute the derivative of ( f(a) ) to see if it's increasing or decreasing.But that might be complicated. Alternatively, perhaps I can test specific values of ( a ) to see if I can find a pattern or a lower bound.Wait, let me try ( a = 2 ):( N = 16 + 8 + 4 + 2 + 1 = 31 )( D = 8 + 4 + 2 + 1 = 15 )So, ( f(2) = 31/15 ≈ 2.0667 )Compare this to ( (5/8)(2 + 1) = (5/8)(3) = 15/8 = 1.875 ). So, ( f(2) ≈ 2.0667 > 1.875 ).Similarly, try ( a = 1/2 ):( N = (1/16) + (1/8) + (1/4) + (1/2) + 1 = (1 + 2 + 4 + 8 + 16)/16 = 31/16 ≈ 1.9375 )( D = (1/8) + (1/4) + (1/2) + 1 = (1 + 2 + 4 + 8)/8 = 15/8 = 1.875 )So, ( f(1/2) = (31/16)/(15/8) = (31/16)*(8/15) = 31/30 ≈ 1.0333 )Compare this to ( (5/8)(1/2 + 1) = (5/8)(3/2) = 15/16 ≈ 0.9375 ). So, ( f(1/2) ≈ 1.0333 > 0.9375 ).Hmm, so in both cases, ( f(a) > (5/8)(a + 1) ). Maybe this is a general inequality.Let me test another value, say ( a = 3 ):( N = 81 + 27 + 9 + 3 + 1 = 121 )( D = 27 + 9 + 3 + 1 = 40 )So, ( f(3) = 121/40 = 3.025 )Compare to ( (5/8)(3 + 1) = (5/8)(4) = 2.5 ). So, ( 3.025 > 2.5 ).Another test: ( a = 1/3 ):( N = (1/81) + (1/27) + (1/9) + (1/3) + 1 ≈ 0.0123 + 0.037 + 0.111 + 0.333 + 1 ≈ 1.493 )( D = (1/27) + (1/9) + (1/3) + 1 ≈ 0.037 + 0.111 + 0.333 + 1 ≈ 1.481 )So, ( f(1/3) ≈ 1.493 / 1.481 ≈ 1.008 )Compare to ( (5/8)(1/3 + 1) = (5/8)(4/3) = 20/24 = 5/6 ≈ 0.833 ). So, ( 1.008 > 0.833 ).Hmm, so in all these test cases, ( f(a) > (5/8)(a + 1) ). Maybe this is always true.If that's the case, then similarly for ( b ), ( frac{b^5 - 1}{b^4 - 1} > frac{5}{8}(b + 1) ).Therefore, multiplying these two inequalities, we would get:left( frac{a^5 - 1}{a^4 - 1} right) cdot left( frac{b^5 - 1}{b^4 - 1} right) > left( frac{5}{8}(a + 1) right) cdot left( frac{5}{8}(b + 1) right) = frac{25}{64}(a + 1)(b + 1)Which is exactly what we need to prove. So, if I can show that ( frac{a^4 + a^3 + a^2 + a + 1}{a^3 + a^2 + a + 1} > frac{5}{8}(a + 1) ), then the original inequality follows.So, let's try to prove this inequality:frac{a^4 + a^3 + a^2 + a + 1}{a^3 + a^2 + a + 1} > frac{5}{8}(a + 1)Multiply both sides by ( a^3 + a^2 + a + 1 ) (which is positive since ( a > 0 )):a^4 + a^3 + a^2 + a + 1 > frac{5}{8}(a + 1)(a^3 + a^2 + a + 1)Let me expand the right-hand side:First, compute ( (a + 1)(a^3 + a^2 + a + 1) ):( (a + 1)(a^3 + a^2 + a + 1) = a^4 + a^3 + a^2 + a + a^3 + a^2 + a + 1 )Combine like terms:- ( a^4 )- ( a^3 + a^3 = 2a^3 )- ( a^2 + a^2 = 2a^2 )- ( a + a = 2a )- ( 1 )So, ( (a + 1)(a^3 + a^2 + a + 1) = a^4 + 2a^3 + 2a^2 + 2a + 1 )Therefore, the inequality becomes:a^4 + a^3 + a^2 + a + 1 > frac{5}{8}(a^4 + 2a^3 + 2a^2 + 2a + 1)Multiply both sides by 8 to eliminate the fraction:8(a^4 + a^3 + a^2 + a + 1) > 5(a^4 + 2a^3 + 2a^2 + 2a + 1)Expand both sides:Left side: ( 8a^4 + 8a^3 + 8a^2 + 8a + 8 )Right side: ( 5a^4 + 10a^3 + 10a^2 + 10a + 5 )Subtract the right side from both sides:8a^4 + 8a^3 + 8a^2 + 8a + 8 - 5a^4 - 10a^3 - 10a^2 - 10a - 5 > 0Simplify:- ( 8a^4 - 5a^4 = 3a^4 )- ( 8a^3 - 10a^3 = -2a^3 )- ( 8a^2 - 10a^2 = -2a^2 )- ( 8a - 10a = -2a )- ( 8 - 5 = 3 )So, the inequality becomes:3a^4 - 2a^3 - 2a^2 - 2a + 3 > 0Hmm, I need to show that this quartic polynomial is positive for all ( a > 0 ), ( a neq 1 ).Let me try to factor this polynomial or find a way to express it as a sum of squares or something similar.Looking at the polynomial ( 3a^4 - 2a^3 - 2a^2 - 2a + 3 ), perhaps I can group terms:Let me group the first and last terms, and the middle terms:( 3a^4 + 3 - 2a^3 - 2a^2 - 2a )Factor out 3 from the first two terms:( 3(a^4 + 1) - 2(a^3 + a^2 + a) )Hmm, not sure if that helps. Alternatively, maybe I can factor it as a product of quadratics.Alternatively, perhaps I can write it as ( (a^2 - 1)^2 + ) something.Let me compute ( (a^2 - 1)^2 = a^4 - 2a^2 + 1 ). Hmm, our polynomial has ( 3a^4 ), so maybe scaling that.Wait, let me try:Let me write ( 3a^4 - 2a^3 - 2a^2 - 2a + 3 ) as:( 3a^4 - 2a^3 - 2a^2 - 2a + 3 = 3(a^4 + 1) - 2(a^3 + a^2 + a) )But I don't see an immediate way to factor this. Maybe I can use the AM-GM inequality or some other inequality.Alternatively, perhaps I can consider that for ( a > 0 ), ( a neq 1 ), the polynomial is always positive.Let me test ( a = 2 ):( 3(16) - 2(8) - 2(4) - 2(2) + 3 = 48 - 16 - 8 - 4 + 3 = 48 - 28 + 3 = 23 > 0 )Good.( a = 1/2 ):( 3(1/16) - 2(1/8) - 2(1/4) - 2(1/2) + 3 = 3/16 - 1/4 - 1/2 - 1 + 3 )Convert to 16ths:( 3/16 - 4/16 - 8/16 - 16/16 + 48/16 = (3 - 4 - 8 - 16 + 48)/16 = (3 - 4= -1; -1 -8= -9; -9 -16= -25; -25 +48=23)/16 = 23/16 > 0 )Good.( a = 1 ): Wait, ( a neq 1 ), but let's see:( 3(1) - 2(1) - 2(1) - 2(1) + 3 = 3 - 2 - 2 - 2 + 3 = (3 + 3) - (2 + 2 + 2) = 6 - 6 = 0 ). So, at ( a = 1 ), it's zero, but ( a neq 1 ), so it's positive around ( a = 1 ).Let me check ( a = 0.5 ) again, which we did, it's positive.What about ( a ) approaching 0? Let's see:As ( a to 0^+ ), the polynomial approaches ( 0 - 0 - 0 - 0 + 3 = 3 > 0 ).As ( a to infty ), the leading term is ( 3a^4 ), which dominates, so it goes to infinity, which is positive.So, the polynomial is positive at ( a = 0 ), positive as ( a to infty ), and positive at ( a = 1 ) (but equals zero). Wait, but at ( a = 1 ), it's zero, but since ( a neq 1 ), we need to ensure that it's positive for all ( a > 0 ), ( a neq 1 ).Wait, but when ( a = 1 ), the original expression is undefined because denominator becomes zero. So, we don't need to consider ( a = 1 ).But just to be thorough, let's check the derivative of the polynomial to see if it has any minima where it could be negative.Let ( P(a) = 3a^4 - 2a^3 - 2a^2 - 2a + 3 ).Compute ( P'(a) = 12a^3 - 6a^2 - 4a - 2 ).Set ( P'(a) = 0 ):( 12a^3 - 6a^2 - 4a - 2 = 0 )This is a cubic equation, which might have one real root. Let me try to find rational roots using Rational Root Theorem. Possible roots are ( pm1, pm2, pm1/2, pm1/3, pm1/4, pm1/6, pm1/12 ).Testing ( a = 1 ):( 12 - 6 - 4 - 2 = 0 ). So, ( a = 1 ) is a root.Therefore, we can factor ( (a - 1) ) out of the derivative:Using polynomial division or synthetic division:Divide ( 12a^3 - 6a^2 - 4a - 2 ) by ( (a - 1) ).Using synthetic division:Coefficients: 12, -6, -4, -2Bring down 12.Multiply by 1: 12.Add to next coefficient: -6 + 12 = 6.Multiply by 1: 6.Add to next coefficient: -4 + 6 = 2.Multiply by 1: 2.Add to last coefficient: -2 + 2 = 0.So, the derivative factors as ( (a - 1)(12a^2 + 6a + 2) ).Thus, ( P'(a) = (a - 1)(12a^2 + 6a + 2) ).Now, ( 12a^2 + 6a + 2 ) is always positive because the discriminant is ( 36 - 96 = -60 < 0 ), so no real roots, and the coefficient of ( a^2 ) is positive.Therefore, the sign of ( P'(a) ) depends on ( (a - 1) ):- For ( a < 1 ), ( (a - 1) < 0 ), so ( P'(a) < 0 ).- For ( a > 1 ), ( (a - 1) > 0 ), so ( P'(a) > 0 ).Thus, ( P(a) ) has a minimum at ( a = 1 ). But ( P(1) = 0 ), as we saw earlier.Therefore, for ( a > 0 ), ( a neq 1 ), ( P(a) > 0 ).Hence, ( 3a^4 - 2a^3 - 2a^2 - 2a + 3 > 0 ) for all ( a > 0 ), ( a neq 1 ).Therefore, the inequality ( frac{a^4 + a^3 + a^2 + a + 1}{a^3 + a^2 + a + 1} > frac{5}{8}(a + 1) ) holds for all ( a > 0 ), ( a neq 1 ).Similarly, the same argument applies to ( b ), so:frac{b^5 - 1}{b^4 - 1} > frac{5}{8}(b + 1)Multiplying these two inequalities together:left( frac{a^5 - 1}{a^4 - 1} right) cdot left( frac{b^5 - 1}{b^4 - 1} right) > left( frac{5}{8}(a + 1) right) cdot left( frac{5}{8}(b + 1) right) = frac{25}{64}(a + 1)(b + 1)Which is exactly what we needed to prove.So, the key steps were:1. Factor the numerator and denominator of each fraction.2. Simplify each fraction to a form that can be compared to a multiple of ( a + 1 ) and ( b + 1 ).3. Show that each simplified fraction is greater than ( frac{5}{8}(a + 1) ) and ( frac{5}{8}(b + 1) ) respectively.4. Multiply these inequalities to get the desired result.5. Verify that the polynomial derived in the process is always positive for ( a > 0 ), ( a neq 1 ), ensuring the inequality holds.This completes the proof.

💡Alright, so I have this geometry problem here. Let me read it again to make sure I understand it properly."In a circle, segment DE is a diameter with DE = 30. Points F and G lie on the circle such that DE, DF, and EF form a triangle. What is the area of the circle traced by the centroid of triangle DEF as F moves on the circumference of the circle excluding the points D and E?"Hmm, okay. So, we have a circle with diameter DE, which is 30 units long. Points F and G are on the circle, but the problem specifically mentions triangle DEF, so I think G might be a typo or maybe not relevant here. Let me focus on triangle DEF.First, since DE is a diameter, the circle has a radius of 15 units because the radius is half the diameter. So, radius r = 15.Now, the triangle DEF is formed by points D, E, and F, where F is moving along the circumference of the circle, except for points D and E. So, as F moves, the triangle DEF changes, and we need to find the area of the circle traced by the centroid of triangle DEF.Okay, so I need to recall what a centroid is. The centroid of a triangle is the point where the three medians intersect. A median is a line segment from a vertex to the midpoint of the opposite side. The centroid divides each median into a ratio of 2:1, with the longer part being closer to the vertex.So, if I can find the path traced by the centroid as F moves around the circle, that should be another circle, and I need to find its area.Let me visualize this. The original circle has diameter DE, so points D and E are fixed on the circle. Point F is moving along the circumference. The centroid H of triangle DEF will move as F moves.Since DE is a diameter, the center of the circle, let's call it O, is the midpoint of DE. So, O is fixed, and it's also the center of the original circle.Now, let me think about the centroid H. Since O is the midpoint of DE, the median from F to DE will pass through O. Wait, is that correct?No, actually, the median from F would go to the midpoint of DE, which is O. So, the median from F is the line segment from F to O. The centroid H is located 1/3 of the way along this median from F towards O.So, as F moves around the circle, the centroid H will trace out a circle that's scaled down by a factor of 1/3 from the original circle.Let me confirm that. If the radius of the original circle is 15, then the distance from O to F is 15. The centroid H is 1/3 of the way from F to O, so the distance from O to H is 15 * (1/3) = 5.Therefore, the centroid H traces a circle with radius 5. The area of this circle would be π * (5)^2 = 25π.Wait, let me make sure I didn't skip any steps. Is the centroid always 1/3 of the way from the vertex to the midpoint of the opposite side? Yes, that's correct. So, since F is moving on a circle of radius 15, the centroid H is moving on a circle of radius 5.So, the area is 25π.I think that makes sense. Let me think if there's another way to approach this problem.Alternatively, I could use coordinate geometry. Let me set up a coordinate system with O at the origin (0,0). Then, points D and E are at (-15, 0) and (15, 0), respectively.Point F is any point on the circle, so its coordinates can be represented as (15 cos θ, 15 sin θ), where θ is the angle parameter.Now, the centroid H of triangle DEF has coordinates that are the average of the coordinates of D, E, and F.So, the x-coordinate of H is (x_D + x_E + x_F)/3 = (-15 + 15 + 15 cos θ)/3 = (0 + 15 cos θ)/3 = 5 cos θ.Similarly, the y-coordinate of H is (y_D + y_E + y_F)/3 = (0 + 0 + 15 sin θ)/3 = 5 sin θ.So, the coordinates of H are (5 cos θ, 5 sin θ), which is the parametric equation of a circle with radius 5 centered at the origin.Therefore, the area of this circle is π*(5)^2 = 25π.Yes, that confirms my earlier reasoning. So, the area is 25π.I think that's solid. I considered both the geometric reasoning about the centroid being 1/3 the distance from F to O and the coordinate geometry approach, which gave the same result. So, I'm confident that 25π is the correct answer.**Final Answer**The area of the circle traced by the centroid is boxed{25pi}.