TITLE: proving :$\frac{(ab+b)(2b+1)}{(ab+a)(5b+1)}+\frac{(bc+c)(2c+1)}{(bc+b)(5c+1)}+\frac{(ca+a)(2a+1)}{(ca+c)(5a+1)}\ge\frac{3}{2}$
QUESTION [7 upvotes]: Let $a,b,c, >0$ be real numbers such that $$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge3$$
How to prove that :
$$\frac{(ab+b)(2b+1)}{(ab+a)(5b+1)}+\frac{(bc+c)(2c+1)}{(bc+b)(5c+1)}+\frac{(ca+a)(2a+1)}{(ca+c)(5a+1)}\ge\frac{3}{2}$$

REPLY [2 votes]: Let $a=\frac{1}{x} ,b=\frac{1}{y} ,c=\frac{1}{z} $
$ \frac{(x+1)(y+2)}{(y+1)(y+5)} + \frac{(y+1)(z+2)}{(z+1)(z+5)}+\frac{(z+1)(x+2)}{(x+1)(x+5)} \geq \frac{3}{2} $
note that $\frac{y+2}{(y+1)(y+5)}\ge \frac{3}{4(y+2)} $ 
then is enough to prove that $\frac{x+1}{y+2} + \frac{y+1}{z+2} + \frac{z+1}{x+2} \geq 2$ 
by Cauchy-Schwarz
$\frac{x+1}{y+2} + \frac{y+1}{z+2} + \frac{z+1}{x+2} =$
$\frac{(x+1)^2}{ (x+1)(y+2) }+\frac{(y+1)^2}{(y+1)(z+2)}+\frac{(z+1)^2}{(z+1)(x+2)}\geq \frac{((x+1)+(y+1)+(z+1))^2}{(x+1)(y+2) +(y+1)(z+2) +(z+1)(x+2) } \geq 2$