Pretty nice inequality

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Pretty nice inequality

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#1 h Oct 16, 2011, 10:14 AM • 2 Y

Y by Adventure10, Mango247

let $a,b,c\in\mathbb{R_+}$ and $abc\geq1$ prove that:

$\frac{a+1}{a^{2}+a+1}+\frac{b+1}{b^{2}+b+1}+\frac{c+1}{c^{2}+c+1}\le2$

This post has been edited 1 time. Last edited by alimathematics, Oct 16, 2011, 4:40 PM

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Dr Sonnhard Graubner

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Dr Sonnhard Graubner

#2 h Oct 16, 2011, 11:11 AM • 3 Y

Y by alimathematics, Adventure10, Mango247

hello, i think it must be
$\frac{a+1}{a^2+a+1}+\frac{b+1}{b^2+b+1}+\frac{c^2+1}{c^2+c+1}\le 2$
we have to proof that
$\frac{a+1}{a^2+a+1}+\frac{b+1}{b^2+b+1}+\frac{c^2+1}{c^2+c+1}-2\le 0$
the left-hand side is equivalent to
$1-{a}^{2}{b}^{2}+a+c+b+abc-{a}^{2}{c}^{2}-{b}^{2}{c}^{2}-{a}^{2}c{b}^{ 2}-2\,{a}^{2}{b}^{2}{c}^{2}-a{b}^{2}{c}^{2}+ab+bc+ca-b{a}^{2}{c}^{2} \le 0$
(the denominator is positive after clearing the denominators)
now we get
$(ab+bc+ca)(1-abc)\le 0$
$1-a^2b^2c^2\le 0$
$abc(1-abc)\le 0$
and since
$a^2b^2+a^2c^2+b^2c^2\geq abc(a+b+c)\geq a+b+c$ we have
$a+b+c-a^2b^2-a^2c^2-b^2c^2\le 0$
Sonnhard.

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sqing

#3 h Jan 24, 2022, 10:37 AM

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Let $a,b,c>0$ and $abc\geq1 .$ Prove that
$\frac{a+1}{a^2+a+k}+\frac{b+1}{b^2+b+k}+\frac{c+1}{c^2+c+k}\leq \frac{6}{k+2}$ Where $k=1,2,3,4.$
$\frac{a+1}{a^2+a+k}+\frac{b+1}{b^2+b+k}+\frac{c+1}{c^2+c+k}\leq \frac{3}{2\sqrt k-1}$ Where $5\leq k\in N^+.$
$\frac{ka+1}{a^2+a+1}+\frac{kb+1}{b^2+b+1}+\frac{kc+1}{c^2+c+1}\leq k+1$ Where $k\in N^+.$

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