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Source: Kyiv City MO 2024 Round 2, Problem 10.2

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#1 h Feb 4, 2024, 11:19 PM • 1 Y

Y by rightways

For any positive real numbers $a, b, c, d$ , prove the following inequality:
$(a^2+b^2)(b^2+c^2)(c^2+d^2)(d^2+a^2) \geq 64abcd|(a-b)(b-c)(c-d)(d-a)|$ Proposed by Anton Trygub

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Victoria_Discalceata1

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Victoria_Discalceata1

#2 h Feb 4, 2024, 11:44 PM • 4 Y

Y by ehuseyinyigit, rstenetbg, ytChen, Sedro

$\left(a^2+b^2\right)\left(c^2+d^2\right)=\left((a-b)^2+2ab\right)\left((c-d)^2+2cd\right)\ge 2\sqrt{(a-b)^2\cdot 2ab}\cdot 2\sqrt{(c-d)^2\cdot 2cd}=8\sqrt{abcd}|(a-b)(c-d)|$

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sqing

#3 h Feb 5, 2024, 3:38 AM

Y by

Nice.

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mihaig

#4 h Feb 5, 2024, 4:51 AM

Y by

Victoria_Discalceata1 wrote:

$\left(a^2+b^2\right)\left(c^2+d^2\right)=\left((a-b)^2+2ab\right)\left((c-d)^2+2cd\right)\ge 2\sqrt{(a-b)^2\cdot 2ab}\cdot 2\sqrt{(c-d)^2\cdot 2cd}=8\sqrt{abcd}|(a-b)(c-d)|$

How did you come up with such a wonderful idea?

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sqing

#5 h Feb 5, 2024, 8:01 AM

Y by

MS_Kekas wrote:

For any positive real numbers $a, b, c, d$ , prove the following inequality:
$(a^2+b^2)(b^2+c^2)(c^2+d^2)(d^2+a^2) \geq 64abcd|(a-b)(b-c)(c-d)(d-a)|$ Proposed by Anton Trygub

Generalization of ytChen :
For any positive real numbers $a, b, c, d$ , prove the following inequality: $\prod_{i=1}^n\left(a_i^2+a_{i+1}^2\right) \ge \left(2\sqrt2\right)^n \prod_{i=1}^na_i\cdot\left| \prod_{i=1}^n(a_i-a_{i+1})\right|,$ where $a_{n+1}=a_1$ .

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#6 h Feb 14, 2024, 1:54 PM • 1 Y

Y by Eka01

sqing wrote:

MS_Kekas wrote:

For any positive real numbers $a, b, c, d$ , prove the following inequality:
$(a^2+b^2)(b^2+c^2)(c^2+d^2)(d^2+a^2) \geq 64abcd|(a-b)(b-c)(c-d)(d-a)|$ Proposed by Anton Trygub

Generalization of ytChen :
For any positive real numbers $a, b, c, d$ , prove the following inequality: $\prod_{i=1}^n\left(a_i^2+a_{i+1}^2\right) \ge \left(2\sqrt2\right)^n \prod_{i=1}^na_i\cdot\left| \prod_{i=1}^n(a_i-a_{i+1})\right|,$ where $a_{n+1}=a_1$ .

Note that
$a_i^2+a_{i+1}^2 = (a_i-a_{i+1})^2+2a_i a_{i+1} \stackrel{\text{AM-GM}}{\geq} 2\sqrt{2a_i a_{i+1}}\cdot\left|a_i-a_{i+1}\right|$ $\implies \prod_{i=1}^n (a_i^2+a_{i+1}^2)\geq \prod_{i=1}^n 2\sqrt{2a_i a_{i+1}}\cdot\left|a_i-a_{i+1}\right| = \prod_{i=1}^n 2\sqrt2 \cdot \prod_{i=1}^n \sqrt{a_ia_{i+1}} \cdot \prod_{i=1}^n \left|a_i-a_{i+1}\right|= \left(2\sqrt2\right)^n \prod_{i=1}^n a_i \cdot \left|\prod_{i=1}^n (a_i-a_{i+1})\right|.\ \square$

This post has been edited 3 times. Last edited by Sammy27, Feb 14, 2024, 4:30 PM

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sqing

#7 h Feb 15, 2024, 1:26 AM

Y by

Thanks.

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