2020年3月14日 · will give products that look like xy3(z4 − 1) x y 3 (z 4 − 1), which give us six degree- 8 8 terms (such as xy3z4 x y 3 z 4) and six degree- 4 4 terms (such as −xy3 − x y 3). We know the determinant is divisible by (x − y)(y − z)(x − z) (x − y) (y − z) (x − z), because if any two of x, y, z x, y, z are equal, the determinant is 0 0. Factoring that out, we should get the ...

Want to improve this question? Add details and clarify the problem by editing this post.

2021年7月16日 · Given non-negative numbers x, y, z such that x + y + z = 3. How do I maximize xy + yz + zx − xyz. I found out that the maximum is 2 and equality holds when one of the numbers is 1 and the other two sum up to 2 but had no idea how to prove it. Can someone help please?

2019年10月14日 · If x, y, z> 0 x, y, z> 0 and x + y + z = 1 x + y + z = 1, prove xy + yz + zx ≥ 9xyz x y + y z + z x ≥ 9 x y z. Can anyone help me? I think I'm close, but I can't get to the end. Using geometric and arithmetic inequality, I got (xy + yz + xz)(2 + xyz) ≥ 9xyz (x y + y z + x z) (2 + x y z) ≥ 9 x y z. Thank you

2015年7月21日 · Prove that xy + yz + zx ≤x2 +y2 +z2 x y + y z + z x ≤ x 2 + y 2 + z 2 . Hint: Use a+b 2 ≥ ab−−√ a + b 2 ≥ a b First I tried using the hint by setting a = x a = x and b = y + z b = y + z, however this results in the inequality:

2019年4月18日 · With a Lagrangian multiplier λ λ in a Lagrangian L:= xy + yz + zx + λ(4 − x − 2y − z) L:= x y + y z + z x + λ (4 − x − 2 y − z), 0 =∂xL = y + z − λ 0 = ∂ x L = y + z − λ etc. gives λ = y + z = z+x 2 = x + y λ = y + z = z + x 2 = x + y. Comparing the first and last of these expressions for λ λ gives x = z x = z, so y + z = z y + z = z and y = 0 y = 0. Then x = 4− ...

Let x = tan 12A x = tan 1 2 A, etc for y, z y, z and B, C B, C. Using the sum of three angles formula, the condition xy + yz + xz = 1 x y + y z + x z = 1 implies that

This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.

2018年9月8日 · Find ∂z ∂x ∂ z ∂ x if xy + yz + zx = 1 x y + y z + z x = 1 I don't understand this question at first. It looks like x, y, z x, y, z are dependent. So we proceed differentiation partially wrt x:

2023年1月10日 · My own solution Consider f(x, y, z) = xy + yz + zx − 2xyz Since f(x, y, z) is symmetric, it will achieve it's maximum value when x = y = z = 1 3 thereby giving 7 27 as the answer. What I want to know, is whether there is any other practical method to solve this question as considering (1 − 2x)(1 − 2y)(1 − 2z) seems somewhat impractical and unimpressive to me. Like how will one think of ...