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Nice inequality

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41195 posts

sqing

#1 h May 17, 2018, 3:25 AM • 2 Y

Y by Adventure10, Mango247

Let $a_1,a_2,\cdots,a_n$ $(n\geq 2)$ be positive real numbers.Prove that $\frac{a^2_1}{a^2_2}+\frac{a^2_2}{a^2_3}+\cdots+\frac{a^2_{n-1}}{a^2_n}+\frac{a^2_n}{a^2_1} \geq \frac{a_1}{a_3}+\frac{a_2}{a_4}+\cdots+\frac{a_{n-1}}{a_1}+\frac{a_n}{a_2}$

This post has been edited 2 times. Last edited by sqing, May 17, 2018, 10:54 AM

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arqady

#2 h May 17, 2018, 4:02 AM • 2 Y

Y by Adventure10, Mango247

It's just Muirhead.

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adhikariprajitraj

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adhikariprajitraj

#3 h May 17, 2018, 4:09 AM • 2 Y

Y by Adventure10, Mango247

arqady wrote:

It's just Muirhead.

But, how can we apply that. Please answer.

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#4 h May 17, 2018, 6:18 AM • 2 Y

Y by Adventure10, Mango247

Multiply both sides bu $\prod_{1}^{n}a_1^2$ and it will be true cause of majorizing [4,2,2,...,0] to [3,2,2,...,1]

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#5 h May 17, 2018, 8:57 AM • 2 Y

Y by Adventure10, Mango247

I found a counter-example for n=4, $a_1=20,a_2=5,a_3=2,a_4=1/2$ .

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sqing

#6 h May 17, 2018, 9:03 AM • 2 Y

Y by Adventure10, Mango247

yongyao wrote:

I found a counter-example for n=4, $a_1=20,a_2=5,a_3=2,a_4=1/2$ .

you are right.
Thank you.
Let $a_1,a_2,\cdots,a_n$ $(2\leq n\leq 4)$ be positive real numbers . Prove that $\frac{a^3_1}{a^3_2}+\frac{a^3_2}{a^3_3}+\cdots+\frac{a^3_{n-1}}{a^3_n}+\frac{a^3_n}{a^3_1} \geq \frac{a_1}{a_n}+\frac{a_2}{a_1}+\frac{a_3}{a_2}+\cdots+\frac{a_n}{a_{n-1}}$

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arqady

#8 h May 17, 2018, 12:23 PM • 2 Y

Y by Adventure10, Mango247

adhikariprajitraj, the problem was changed.

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41195 posts

sqing

#10 h Jun 3, 2018, 1:54 AM • 2 Y

Y by Adventure10, Mango247

sqing wrote:

Let $a_1,a_2,\cdots,a_n$ $(n\geq 2)$ be positive real numbers.Prove that $\frac{a^2_1}{a^2_2}+\frac{a^2_2}{a^2_3}+\cdots+\frac{a^2_{n-1}}{a^2_n}+\frac{a^2_n}{a^2_1} \geq \frac{a_1}{a_3}+\frac{a_2}{a_4}+\cdots+\frac{a_{n-1}}{a_1}+\frac{a_n}{a_2}$

Nordic Mathematical Contest 1987:
Let $a, b$ , and $c$ be positive real numbers.Prove that $\frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2}\ge \frac{a}{b}+ \frac{b}{c}+ \frac{c}{a}$ here

This post has been edited 2 times. Last edited by sqing, Jun 3, 2018, 2:51 AM

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