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Nice inequality

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Cauchy Inequality

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#1 h Jan 20, 2014, 10:02 AM • 2 Y

Y by Adventure10, Mango247

Let $a,b,c$ be positive real numbers such $\frac{1}{a^2+b^2+1}+\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1} \geq 1$
Prove that $ab+bc+ca \leq 3.$

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ISHO95

#2 h Jan 20, 2014, 11:48 AM • 2 Y

Y by Adventure10, Mango247

From cauchy inequality we get
$\frac{1}{a^2+b^2+1}\le\frac{c^2+2}{(a+b+c)^2}$
then
$\frac{a^2+b^2+c^2+6}{(a+b+c)^2}\ge 1$
or
$ab+bc+ca \le 3$
done!

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sqing

#3 h Jul 30, 2015, 9:09 AM • 2 Y

Y by Adventure10, Mango247

Let $a,b,c$ be positive real numbers such $\frac{1}{a^2+b^2+1}+\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1} \geq 1$ .Prove that $a^2b^2+b^2c^2+c^2a^2 \leq 3.$

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arqady

#4 h Aug 2, 2015, 9:53 AM • 2 Y

Y by Adventure10, Mango247

sqing wrote:

Let $a,b,c$ be positive real numbers such $\frac{1}{a^2+b^2+1}+\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1} \geq 1$ .Prove that $a^2b^2+b^2c^2+c^2a^2 \leq 3.$

We need to prove that $ab+ac+bc\leq3$ for all positives $a$ , $b$ and $c$ such that $\sum_{cyc}\frac{1}{a+b+c}\geq1$ .
Let $ab+ac+bc>3$ .
The condition gives $\prod_{cyc}(a+b)\leq2+2(a+b+c)$ .
But $2+2(a+b+c)<\frac{2}{3}(a+b+c)(ab+ac+bc)+2\sqrt{\frac{(ab+ac+bc)^3}{27}}$ .
We'll prove that $\frac{2}{3}(a+b+c)(ab+ac+bc)+2\sqrt{\frac{(ab+ac+bc)^3}{27}}\leq\prod_{cyc}(a+b)$ .
Let $a+b+c=3u$ , $ab+ac+bc=3v^2$ , where $v>0$ and $abc=w^3$ .
Hence, we need to prove that $6uv^2+2v^3\leq9uv^2-w^3$ , which is $f(w^3)\leq0$ , where $f$ is an increasing function.
Thus, $f$ gets a maximal value for maximal value of $w^3$ , which happens for equality of two variables.
Since the inequality $f(w^3)\leq0$ is homogeneous, it remains to prove the last inequality for $b=c=1$ ,
which gives $(a-1)^2(3a^2+4a+2)\geq0$ .
Id est, we have obtained a contradiction. Done!

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