Inequality

Let x,y,z be +ve reals such that xy+yz+zx=5. How to Prove that 3x^2+3y^2+z^2>=10???

#Algebra

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Comments

Okay, since my skills are as rotten as blue cheese, I'll try my best.

We have that $\begin{aligned} xy + yz + zx &= 3 \\ \sqrt 3(xy + yz + zx) &= 3\sqrt 3\\ z(\sqrt 3x + \sqrt 3y) &= \sqrt 3(3 - xy)\\ \sqrt 3x + \sqrt 3y &= \dfrac{\sqrt 3(3 - xy)}{z}\\ (\sqrt 3x + \sqrt 3y)^2 &= \dfrac{3(3 - xy)^2}{z^2} \end{aligned}$ and $3x^2 + 3y^2 \ge \dfrac{(\sqrt 3x + \sqrt 3y)^2}{2}$ so $\begin{aligned} 3x^2 + 3y^2 &\ge \dfrac{3(3 - xy)^2}{2z^2}\\ 3x^2 + 3y^2 + z^2 &\ge \dfrac{3(3 - xy)^2}{2z^2} + z^2 \end{aligned}$ .

And I gave up. Could anybody give me a hand, please?

Thành Đạt Lê - 2 years, 5 months ago

This is false. For $(x,y,z) = \left( \sqrt{ \frac35 }, \sqrt{ \frac35 } , 2\sqrt{ \frac35 } \right)$ , we have $3x^2 + 3y^2 + z^2 = 6 \not \geq 10$ .

Pi Han Goh - 2 years, 5 months ago

@Pi Han Goh if xy+yz+zx=5 then what is the answer?

Raj Mantri - 2 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$