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Suppose $x,y,z$ are real numbers. Prove that $yz+zx+xy\leq (y+z-x)^2+(z+x-y)^2+(x+y-z)^2$.

May I ask how to use Cauchy-Schwarz Inequality to obtain $|yz+zx+xy|\leq x^2+y^2+z^2$ ? I only know $x^2+y^2+z^2\geq yz+zx+xy $. How to deal with the absolute sign?

To finish the proof, I think I need to check whether $++$ is negative or not because $(y+z-x)^2+(z+x-y)^2+(x+y-z)^2$ must be positive. Am I right? But how to check that? Or are there any easier ways to solve the problem(also with the use of Cauchy-Schwarz Inequality) ? Thank you.

sunny
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    Just by expanding the right side you can get a proof easily. And also $|yz+zx+xy| \le |y||z|+|z||x|+|x||y| \le |x|^2+|y|^2+|z|^2=x^2+y^2+z^2$. – Zerox Mar 27 '22 at 05:36

1 Answers1

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Let us simplify the RHS first..

$(y + z - x)^2 = x^2 + y^2 + z^2 + 2 ( y z - x y - x z)$

$(z + x - y)^2 = x^2 + y^2 + z^2 + 2 (x z - x y - y z)$

$(x + y - z)^2 = x^2 + y^2 + z^2 + 2 (x y - x z - y z )$

Adding the above equations, we get

$(y + z - x)^2 + (z + x - y)^2 + (x + y - z)^2 = 3 (x^2 + y^2 + z^2) - 2 (x y + y z + x z )$.

Thus, it follows that

$(y + z - x)^2 + (z + x - y)^2 + (x + y - z)^2 - (x y + y z + z x) = A$ $ \ \ \ \ \ $ (1)

Here,

$A = 3 (x^2 + y^2 + z^2) - 3 (x y + y z + z x) $

i.e.

$A = 3 [ x^2 + y^2 + z^2 - (x y + y z + z x) ]$

By Cauchy-Schwarz inequality, it is immediate that

$ | x y + y z + z x | \leq x^2 + y^2 + z^2 $.

Thus, we deduce that

$ (x y + y z + z x) \leq | x y + y z + z x | \leq x^2 + y^2 + z^2 $.

This shows that

$A = 3 [ x^2 + y^2 + z^2 - (x y + y z + z x) ] \geq 0$. $ \ \ \ \ \ $ (2)

Using (1) and (2), we conclude that

$ x y + y z + z x \leq (y + z - x)^2 + (z + x - y)^2 + (x + y - z)^2$

for all real numbers $x, y, z$.

This completes the proof. $\blacksquare$

Dr. Sundar
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