Mmo 9-10 graders P5

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Mmo 9-10 graders P5

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153 posts

Bet667

#1 h Apr 3, 2025, 6:54 AM • 1 Y

Y by cubres

Let $a,b,c,d$ be real numbers less than 2.Then prove that $\frac{a^3}{b^2+4}+\frac{b^3}{c^2+4}+\frac{c^3}{d^2+4}+\frac{d^3}{a^2+4}\le4$

This post has been edited 2 times. Last edited by Bet667, Apr 3, 2025, 6:56 AM
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Bet667

#2 h Apr 3, 2025, 7:05 AM • 1 Y

Y by cubres

Guyss plss give me a idea or solution

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Bet667

#3 h Apr 3, 2025, 7:37 AM • 1 Y

Y by cubres

Anyonee please

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Quantum-Phantom

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Quantum-Phantom

#4 h Apr 3, 2025, 8:38 AM • 3 Y

Y by DensSv, kiyoras_2001, cubres

By the adjustment method, we can prove that it suffices to prove this inequality for non-negative variables. Now let $s\le t\le u\le v$ be an arrangement of $a$ , $b$ , $c$ , $d$ , then by the rearrangement inequality, we have
$\frac{a^3}{b^2+4}+\frac{b^3}{c^2+4}+\frac{c^3}{d^2+4}+\frac{d^3}{a^2+4}\le\frac{s^3}{v^2+4}+\frac{t^3}{u^2+4}+\frac{u^3}{t^2+4}+\frac{v^3}{s^2+4}.$ Since $s$ , $v\le2$ , then
$\frac{s^3}{v^2+4}+\frac{v^3}{s^2+4}\le\frac{s^3}{v^2+s^2}+\frac{v^3}{s^2+v^2}=\frac{s^3+v^3}{s^2+v^2}\le\frac{2s^2+2v^2}{s^2+v^2}=2.$ Similarly we get $\frac{t^3}{u^2+4}+\frac{u^3}{t^2+4}\le2$ , done.

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Bet667

#5 h Apr 3, 2025, 1:24 PM • 1 Y

Y by cubres

Does anyone has another solution

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