Problem 27: Inequality by Vasile Cirtoaje (2005)

Posts: 312
Joined: Mar 10, 2015

by henderson, Jun 23, 2016, 1:52 PM

$\bf\color{red}Problem \ 27$ Let $a,b,c$ be non-negative real numbers such that $ab+bc+ca=3.$ Prove that
$\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+b^2}\geq \frac{3}{2}.$ $\bf\color{red}My \ solution$ $\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+b^2}\geq \frac{3}{2}$ $\color{blue}\iff$
$\frac{3+2(a^2+b^2+c^2)+(a^2b^2+b^2c^2+c^2a^2)}{(1+a^2)(1+b^2)(1+c^2)}\geq \frac{3}{2}$ $\color{blue}\iff$
$6+4(a^2+b^2+c^2)+2(a^2b^2+b^2c^2+c^2a^2)\geq 3+3(a^2+b^2+c^2)+3(a^2b^2+b^2c^2+c^2a^2)+3a^2b^2c^2$ $\color{blue}\iff$
$\color{blue}3+(a^2+b^2+c^2)\geq (a^2b^2+b^2c^2+c^2a^2)+3a^2b^2c^2.$ Now, let's make the inequality homogenous using $ab+bc+ca=3:$
$\color{red}(ab+bc+ca)^3+(ab+bc+ca)^2(a^2+b^2+c^2)\geq 3(ab+bc+ca)(a^2b^2+b^2c^2+c^2a^2)+27a^2b^2c^2$ $\color{blue}\iff$
$\color{red}a^3b^3+b^3c^3+c^3a^3+3(ab+bc)(bc+ca)(ca+ab)+(a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c))(a^2+b^2+c^2)\geq 3(a^3b^3+b^3c^3+c^3a^3)+3abc(a^2b+b^2c+c^2a)+3abc(a^2c+b^2a+c^2b)+27a^2b^2c^2$ $\color{blue}\iff$
$\color{red}a^3b^3+b^3c^3+c^3a^3+3abc(a^2c+b^2a+c^2b)+3abc(a^2b+b^2c+c^2a)+6a^2b^2c^2+(a^4b^2+b^4c^2+c^4a^2)+(a^4c^2+b^4a^2+c^4b^2)+3a^2b^2c^2+2abc(a^3+b^3+c^3)+2abc(a^2b+b^2c+c^2a)+2abc(a^2c+b^2a+c^2b)\geq 3(a^3b^3+b^3c^3+c^3a^3)+3abc(a^2b+b^2c+c^2a)+3abc(a^2c+b^2a+c^2b)+27a^2b^2c^2$
$\color{blue}\iff$
$\color{red}(a^4b^2+b^4c^2+c^4a^2)+(a^4c^2+b^4a^2+c^4b^2)+2abc(a^3+b^3+c^3)+2abc(a^2b+b^2c+c^2a)+2abc(a^2c+b^2a+c^2b)\geq 2(a^3b^3+b^3c^3+c^3a^3)+18a^2b^2c^2.$
But, the last inequality is obvious since
$(a^4b^2+b^4c^2+c^4a^2)+(a^4c^2+b^4a^2+c^4b^2)\geq 2(a^3b^3+b^3c^3+c^3a^3)$ and
$2abc(a^3+b^3+c^3)+2abc(a^2b+b^2c+c^2a)+2abc(a^2c+b^2a+c^2b)\geq 18a^2b^2c^2.$ We are done!