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JK1603JK   4
N 17 minutes ago by Quantum-Phantom
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Let a,b,c>=0: ab+bc+ca=3 then maximize P=\frac{a^2b+b^2c+c^2a+9}{a+b+c}+\frac{abc}{2}.
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inequality
senku23   3
N Wednesday at 2:02 PM by SunnyEvan
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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senku23
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Wednesday at 2:02 PM
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senku23
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#1 h  Wednesday at 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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giangtruong13
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giangtruong13
#2 h  Wednesday at 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, Wednesday at 11:24 AM
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User21837561
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User21837561
#3 h  Wednesday at 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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SunnyEvan
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SunnyEvan
#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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