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#1 h Mar 19, 2025, 11:17 AM

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Let x,y,z in R+ prove that 8(x^3+y^3+z^3)2≥9(x^2+yz)(y^2+zx)(z^2+xy).

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#2 h Mar 19, 2025, 11:24 AM

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senku23 wrote:

Let x,y,z in R+ prove that 8(x^3+y^3+z^3)2≥9(x^2+yz)(y^2+zx)(z^2+xy).

LaTex: Let $x,y,z$ be real positive numbers. Prove that: $8(x^3+y^3+z^3) \geq 9(x^2+yz)(y^2+xz)(z^2+xy)$

This post has been edited 1 time. Last edited by giangtruong13, Mar 19, 2025, 11:24 AM

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#3 h Mar 19, 2025, 12:47 PM

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Let $x,y,z \in R^+$ . Prove that $8(x^3+y^3+z^3)^2\ge9(x^2+yz)(y^2+zx)(z^2+xy)$ .

Expanding:
$8\sum x^6+16\sum x^3y^3\ge 9(2x^2y^2z^2+\sum x^4yz+\sum x^3y^3)$
$8\sum x^6+7\sum x^3y^3\ge 9\sum x^4yz+18x^2y^2z^2$

Since $\sum x^6+\sum x^3y^3\ge\sum x^6+\sum x^2y^2z^2\ge\sum x^4(y^2+z^2)\ge2\sum x^4yz$
by AM-GM and Schur,

$8\sum x^6+7\sum x^3y^3\ge 7\sum x^6+6\sum x^3y^3+2\sum x^4yz\ge 9\sum x^4yz + 18x^2y^2z^2$ by AM-GM.

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#5 h Wednesday at 2:02 PM

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Let $x,y,z \in R^+$ . Prove that $8(x^3+y^3+z^3)^2\geq 9(x^2+yz)(y^2+zx)(z^2+xy)$ .
Holder : $8(x^3+y^3+z^3)^2\geq \frac{8}{3}(\sum x^2)^3$ AM-GM : $(\sum (x^2+yz))^3 \geq 27(x^2+yz)(y^2+zx)(z^2+xy)$ $8(\sum x^2)^3 \geq (\sum (x^2+yz))^3$ done .

This post has been edited 3 times. Last edited by SunnyEvan, Wednesday at 2:04 PM

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