I'm having trouble solving the following inequality problem:
Show that: $\forall$ $x,y,z > 0$ : $$\sum_{cyc}\frac{x}{xz+2x+1} \leq \frac{3}{4} $$ Here's my take :
First , i used the symmetry of the $LHS$ and assumed WLOG: $x\geq y \geq z$ .
Furthermore i get the three following inequalities based on our first assumption:
$xz+2x+1\geq z^2+2z+1 = (z+1)^2$
$xy+2y+1\geq z^2+2z+1 = (z+1)^2$
$zy+2z+1\geq z^2+2z+1 = (z+1)^2$
Therefore:
$$LHS \leq \frac{x+y+z}{(z+1)^2}$$
Now , From AM-GM we have : $z+1 \geq 2\sqrt{z} \implies (z+1)^2 \geq 4z \implies \frac{1}{(z+1)^2} \leq \frac{1}{4z}$
And from the assumption in the beginning : $x+y+z \leq 3x$
Combining the two gives us :
$LHS \leq \frac{3}{4} \cdot \frac{x}{z}$
But then since $x\geq z \hspace{2mm}$ i only get a really loose bound.
I've made many more attemps on the problem but couldn't get anywhere.
I'd like to know where i've failed in my answer and know if there's an answer using well known inequalities (AM-GM , CS , Chebyshev , Jensen's ...).
Thanks in advance for any help given.
