Can you help me this Inequality

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#1 h Mar 10, 2024, 4:26 PM

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Let a,b,c be positive real numbers satisfying: $a+b+c=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}. Prove that :\frac{1}{a^{2}+b^{2}+1}+\frac{1}{b^{2}+c^{2}+1}+\frac{1}{c^ {2}+a^{2}+1}\leq 1$ .

This post has been edited 5 times. Last edited by firstVMO, Mar 10, 2024, 4:30 PM
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#2 h Mar 10, 2024, 5:04 PM

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Does anyone have any ideas?

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arqady

#3 h Mar 10, 2024, 7:27 PM • 2 Y

Y by ehuseyinyigit, firstVMO

firstVMO wrote:

Let a,b,c be positive real numbers satisfying: $a+b+c=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}. Prove that :\frac{1}{a^{2}+b^{2}+1}+\frac{1}{b^{2}+c^{2}+1}+\frac{1}{c^ {2}+a^{2}+1}\leq 1$ .

We need to prove that $\sum_{cyc}\frac{a^2+b^2}{a^2+b^2+1}\geq2.$ Now, by C-S $\sum_{cyc}\frac{a^2+b^2}{a^2+b^2+1}\geq\frac{4(a^2+b^2+c^2)^2}{\sum\limits_{cyc}\left((a^2+b^2)^2+a^2+b^2\right)}$ and it's enough to prove that:
$a^2b^2+a^2c^2+b^2c^2\geq a^2+b^2+c^2$ or
$(a^2b^2+a^2c^2+b^2c^2)\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\geq(a^2+b^2+c^2)(a+b+c)^2$ or $(a^2+b^2+c^2)\left(a^2b+b^2c+c^2a\right)^2\geq(a^2b^2+a^2c^2+b^2c^2)(ab+ac+bc)^2,$ which is trivial.

This post has been edited 2 times. Last edited by arqady, Mar 10, 2024, 7:31 PM

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#5 h Mar 11, 2024, 3:44 AM

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arqady wrote:

firstVMO wrote:

Let a,b,c be positive real numbers satisfying: $a+b+c=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}. Prove that :\frac{1}{a^{2}+b^{2}+1}+\frac{1}{b^{2}+c^{2}+1}+\frac{1}{c^ {2}+a^{2}+1}\leq 1$ .

We need to prove that $\sum_{cyc}\frac{a^2+b^2}{a^2+b^2+1}\geq2.$ Now, by C-S $\sum_{cyc}\frac{a^2+b^2}{a^2+b^2+1}\geq\frac{4(a^2+b^2+c^2)^2}{\sum\limits_{cyc}\left((a^2+b^2)^2+a^2+b^2\right)}$ and it's enough to prove that:
$a^2b^2+a^2c^2+b^2c^2\geq a^2+b^2+c^2$ or
$(a^2b^2+a^2c^2+b^2c^2)\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\geq(a^2+b^2+c^2)(a+b+c)^2$ or $(a^2+b^2+c^2)\left(a^2b+b^2c+c^2a\right)^2\geq(a^2b^2+a^2c^2+b^2c^2)(ab+ac+bc)^2,$ which is trivial.

Thank you

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mudok

#6 h Mar 11, 2024, 7:56 AM • 3 Y

Y by anhduy98, firstVMO, arqady

By Cauchy $(a^2+b^2+1)(1+1+c^2)\ge (a+b+c)^2$

$\sum \frac{1}{a^2+b^2+1}\le \sum \frac{c^2+2}{(a+b+c)^2}$

It suffices to prove

$(a+b+c)^2\ge a^2+b^2+c^2+6 \iff ab+bc+ca\ge 3$

By the condition and Cauchy

$a+b+c=\frac a b +\frac b c +\frac c a \ge \frac{(a+b+c)^2}{ab+bc+ca}$

$\Rightarrow ab+bc+ca\ge a+b+c$

$\Rightarrow (ab+bc+ca)^2\ge (a+b+c)^2\ge 3(ab+bc+ca)$

$\Rightarrow ab+bc+ca\ge 3$

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#7 h Mar 11, 2024, 3:50 PM

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mudok wrote:

By Cauchy $(a^2+b^2+1)(1+1+c^2)\ge (a+b+c)^2$

$\sum \frac{1}{a^2+b^2+1}\le \sum \frac{c^2+2}{(a+b+c)^2}$

It suffices to prove

$(a+b+c)^2\ge a^2+b^2+c^2+6 \iff ab+bc+ca\ge 3$

By the condition and Cauchy

$a+b+c=\frac a b +\frac b c +\frac c a \ge \frac{(a+b+c)^2}{ab+bc+ca}$

$\Rightarrow ab+bc+ca\ge a+b+c$

$\Rightarrow (ab+bc+ca)^2\ge (a+b+c)^2\ge 3(ab+bc+ca)$

$\Rightarrow ab+bc+ca\ge 3$

Thank you.

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