💡Okay, so I have this inequality to prove: for real numbers a, b, and c, the sum of their squares is greater than or equal to one-third of the square of their sum. In symbols, that's a² + b² + c² ≥ (1/3)(a + b + c)². Hmm, I think I remember something about inequalities and maybe the Cauchy-Schwarz inequality or something like that. But let me try to work it out step by step.First, maybe I should expand the right-hand side to see what it looks like. So, (a + b + c)² is a² + b² + c² + 2ab + 2ac + 2bc. Then, if I multiply that by 1/3, I get (1/3)a² + (1/3)b² + (1/3)c² + (2/3)ab + (2/3)ac + (2/3)bc. So the inequality becomes a² + b² + c² ≥ (1/3)a² + (1/3)b² + (1/3)c² + (2/3)ab + (2/3)ac + (2/3)bc.Hmm, maybe I can subtract the right-hand side from both sides to simplify. Let's see: a² + b² + c² - [(1/3)a² + (1/3)b² + (1/3)c² + (2/3)ab + (2/3)ac + (2/3)bc] ≥ 0. Simplifying the left side, I get (2/3)a² + (2/3)b² + (2/3)c² - (2/3)ab - (2/3)ac - (2/3)bc ≥ 0.I can factor out a (2/3) from all the terms: (2/3)(a² + b² + c² - ab - ac - bc) ≥ 0. Since 2/3 is positive, this inequality is equivalent to a² + b² + c² - ab - ac - bc ≥ 0.Now, I need to show that a² + b² + c² - ab - ac - bc is always non-negative for real numbers a, b, c. Maybe I can rewrite this expression in a different form. I recall that sometimes expressions like this can be written as sums of squares, which are always non-negative.Let me try to express a² + b² + c² - ab - ac - bc as a sum of squares. One approach is to complete the square or use some identity. Let's see:I know that (a - b)² = a² - 2ab + b², and similarly for (a - c)² and (b - c)². Maybe if I add these up:(a - b)² + (a - c)² + (b - c)² = (a² - 2ab + b²) + (a² - 2ac + c²) + (b² - 2bc + c²) = 2a² + 2b² + 2c² - 2ab - 2ac - 2bc.Hmm, that's twice the expression I have. So, (a - b)² + (a - c)² + (b - c)² = 2(a² + b² + c² - ab - ac - bc). Therefore, a² + b² + c² - ab - ac - bc = [(a - b)² + (a - c)² + (b - c)²]/2.Since squares are always non-negative, the sum of squares is also non-negative, and dividing by 2 doesn't change that. So, a² + b² + c² - ab - ac - bc is indeed non-negative. Therefore, the original inequality holds.Wait, let me double-check that step where I expressed the quadratic form as a sum of squares. I think that's correct because each squared term is non-negative, so their sum is non-negative, and hence the entire expression is non-negative.Another way to think about this might be using the concept of variance in statistics. If I consider a, b, c as data points, the variance is always non-negative, and it's related to the expression we have. But maybe that's going off track.Alternatively, I could use the Cauchy-Schwarz inequality. The Cauchy-Schwarz inequality states that for vectors u and v, (u·v)² ≤ (u·u)(v·v). If I let u = (a, b, c) and v = (1, 1, 1), then u·v = a + b + c, and u·u = a² + b² + c², v·v = 1 + 1 + 1 = 3.Applying Cauchy-Schwarz: (a + b + c)² ≤ 3(a² + b² + c²). Rearranging this gives a² + b² + c² ≥ (1/3)(a + b + c)², which is exactly what we wanted to prove. So that's another way to see it using a well-known inequality.But I think the first method, expressing the difference as a sum of squares, is more elementary and might be preferred if someone isn't familiar with Cauchy-Schwarz. Either way, both methods confirm the inequality holds.Just to make sure, let me test it with some numbers. Suppose a = b = c = 1. Then the left side is 1 + 1 + 1 = 3, and the right side is (1/3)(3)² = (1/3)(9) = 3. So 3 ≥ 3, which holds.Another example: a = 2, b = 3, c = 4. Left side: 4 + 9 + 16 = 29. Right side: (1/3)(2 + 3 + 4)² = (1/3)(9)² = (1/3)(81) = 27. So 29 ≥ 27, which is true.What if one of them is negative? Let's say a = 1, b = -1, c = 0. Left side: 1 + 1 + 0 = 2. Right side: (1/3)(1 - 1 + 0)² = (1/3)(0)² = 0. So 2 ≥ 0, which is true.Another test: a = 0, b = 0, c = 0. Both sides are 0, so equality holds.What about a = 1, b = 2, c = -3. Left side: 1 + 4 + 9 = 14. Right side: (1/3)(1 + 2 - 3)² = (1/3)(0)² = 0. So 14 ≥ 0, which is true.It seems like the inequality holds in these test cases, which gives me more confidence in the proof.I think I've covered the main approaches: expanding and simplifying, expressing as a sum of squares, using Cauchy-Schwarz, and testing with examples. All of these support the conclusion that the inequality is indeed true for all real numbers a, b, c.

💡Okay, so I have this problem where I'm given two vectors, vector a which is (2,1) and vector b which is (m,2). The question says that vector a is parallel to vector b, and I need to find the value of m. Alright, let's try to figure this out step by step.First, I remember that if two vectors are parallel, one must be a scalar multiple of the other. That means there exists some scalar, let's call it k, such that vector a equals k times vector b, or vice versa. So, in equation form, that would be:(2,1) = k*(m,2)Which means that each component of vector a is equal to k times the corresponding component of vector b. So, breaking that down, we have two equations:2 = k*m ...(1)1 = k*2 ...(2)Now, I can solve equation (2) first to find the value of k. If 1 = k*2, then k must be 1/2. That makes sense because 2 times 1/2 is 1. So, k = 1/2.Now that I have k, I can substitute it back into equation (1) to find m. So, plugging in k = 1/2 into equation (1):2 = (1/2)*mTo solve for m, I can multiply both sides of the equation by 2:2*2 = mSo, m = 4.Wait, let me double-check that. If m is 4, then vector b becomes (4,2). Is (2,1) parallel to (4,2)? Well, if I multiply vector a by 2, I get (4,2), which is exactly vector b. So, yes, they are scalar multiples of each other, which confirms that they are parallel.Hmm, is there another way to check this? Maybe by looking at the slopes if I consider these vectors as lines in a plane. The slope of vector a would be rise over run, which is 1/2. The slope of vector b would be 2/m. If they are parallel, their slopes should be equal. So:1/2 = 2/mSolving for m, cross-multiplying:1*m = 2*2m = 4Yep, that gives the same result. So, m is definitely 4.I think I've got it. The key idea here is that parallel vectors must be scalar multiples of each other, so setting up the proportion between their components allows me to solve for the unknown variable m. Both methods I used—the scalar multiple approach and the slope approach—lead me to the same answer, which gives me confidence that m is indeed 4.

💡Alright, so I have this problem here where I need to find the ordered pair (s, h) such that the parameterization of the line y = 3x - 11 is given by the vector equation:[begin{pmatrix} x y end{pmatrix} = begin{pmatrix} s -2 end{pmatrix} + t begin{pmatrix} 5 h end{pmatrix}]Hmm, okay. Let me try to break this down step by step. I remember that a parameterization of a line involves a point on the line and a direction vector. So, in this case, the point is (s, -2) and the direction vector is (5, h). First, I should verify that the point (s, -2) lies on the line y = 3x - 11. Because if it's a point on the line, it should satisfy the equation. So, plugging x = s and y = -2 into the equation, we get:-2 = 3s - 11Okay, so I can solve for s here. Let me do that:-2 = 3s - 11 Add 11 to both sides: 9 = 3s Divide both sides by 3: s = 3Alright, so s is 3. That wasn't too bad. Now, I need to find h. Hmm, how do I do that? Well, I know that the direction vector (5, h) should be parallel to the line y = 3x - 11. The slope of the line is 3, so the direction vector should have a slope of 3 as well.Wait, the slope of the direction vector is rise over run, which would be h/5. Since the slope of the line is 3, we can set up the equation:h / 5 = 3Solving for h:h = 3 * 5 h = 15Okay, so h is 15. Let me just double-check that. If the direction vector is (5, 15), then the slope is 15/5 = 3, which matches the slope of the line. That makes sense.Just to be thorough, let me check another point on the parameterization. If I set t = 1, then the point would be:x = s + 5t = 3 + 5(1) = 8 y = -2 + h*t = -2 + 15(1) = 13Now, plugging x = 8 into the line equation y = 3x - 11:y = 3*8 - 11 = 24 - 11 = 13Yep, that matches. So, the point (8, 13) is on the line, which confirms that h = 15 is correct.So, putting it all together, s is 3 and h is 15. Therefore, the ordered pair (s, h) is (3, 15).**Final Answer**The ordered pair is boxed{(3, 15)}.

💡Okay, so I have this problem about a geometric sequence. Let me try to understand what it's asking. It says that there's a geometric sequence of positive integers, and the first term is 5. The fourth term is 320. I need to find the third term of this sequence. Hmm, okay.First, I remember that in a geometric sequence, each term is found by multiplying the previous term by a common ratio, which is usually denoted as 'r'. So, if the first term is 5, then the second term would be 5 times r, the third term would be 5 times r squared, and the fourth term would be 5 times r cubed. That makes sense because each time you're multiplying by r again.So, the first term is 5, the second term is 5r, the third term is 5r², and the fourth term is 5r³. Got it. Now, the problem tells me that the fourth term is 320. So, I can set up an equation: 5r³ = 320. My goal is to find the third term, which is 5r², so I need to find the value of r first.Alright, let's solve for r. Starting with 5r³ = 320. To isolate r³, I can divide both sides of the equation by 5. So, r³ = 320 / 5. Let me calculate that. 320 divided by 5 is 64. So, r³ = 64.Now, to find r, I need to take the cube root of both sides. The cube root of 64 is 4 because 4 times 4 times 4 is 64. So, r = 4. Okay, that seems straightforward.Now that I know r is 4, I can find the third term, which is 5r². Let's compute that. First, calculate r squared: 4 squared is 16. Then, multiply that by 5: 5 times 16 is 80. So, the third term should be 80.Wait, let me double-check to make sure I didn't make any mistakes. Starting with the first term as 5, multiplying by r=4 each time: 5, then 5*4=20, then 20*4=80, and then 80*4=320. Yep, that matches the fourth term given in the problem. So, the third term is indeed 80.I think that's it. I didn't encounter any problems in my calculations, and the steps make sense. I used the properties of geometric sequences correctly, solved for the common ratio, and then found the required term. Everything checks out.

💡First, determine James's current age. Since James will be 44 years old after 5 years, subtract 5 from 44:44 - 5 = 39 years oldNext, find Jessica's current age. James is 7 years older than Jessica, so subtract 7 from James's current age:39 - 7 = 32 years oldFinally, calculate Justin's current age. When Justin was born, Jessica was 6 years old, so subtract 6 from Jessica's current age:32 - 6 = 26 years oldJustin is currently 26 years old.

💡First, I need to determine the initial amount of money Jiho had, which I'll denote as ( M ).At the fish shop, Jiho spent half of his money, so the remaining amount after this purchase is ( frac{1}{2}M ).Next, at the fruit shop, Jiho spent an additional 5,000 won on top of one-third of the remaining money from the fish shop. This means he spent ( frac{1}{3} times frac{1}{2}M + 5,000 ) won.After these expenditures, Jiho is left with 5,000 won. I can set up the equation:[frac{1}{2}M - left( frac{1}{3} times frac{1}{2}M + 5,000 right) = 5,000]Simplifying the equation:[frac{1}{2}M - frac{1}{6}M - 5,000 = 5,000]Combining like terms:[frac{2}{6}M = 10,000]Simplifying further:[frac{1}{3}M = 10,000]Finally, solving for ( M ):[M = 10,000 times 3 = 30,000]Therefore, Jiho initially had 30,000 won.