Find max value of $a$ Let $a, x, y, z \in \mathbb{R}_{>0}$ such that $xyz=1$.
Find maximum value of $a$ such that is satisfied the inequality
$$ \sum_{cyc} \frac{x}{(x+1)(y+a)} \ge \frac{3}{2(1+a)} $$
After doing some calculation the max value of $a$ seems to me to be $2$ but I can’t prove it.
 A: Let $y=z$ and $x\rightarrow+\infty.$
Thus, $$\frac{1}{a}\geq\frac{3}{2(1+a)},$$
which gives $a\leq2$.
We'll prove that $a=2$ it's a maximal value.
Indeed, we need to prove that
$$\sum_{cyc}\frac{x}{(x+1)(y+2)}\geq\frac{1}{2}$$ or
$$\sum_{cyc}(2x^2y+xy-3x)\geq0.$$
Now, let $x=\frac{p}{q}$ and $y=\frac{q}{r}$, where $p$, $q$ and $r$ are positives.
Thus, $z=\frac{r}{p}$ and we need to prove that
$$\sum_{cyc}\left(\frac{2p^2}{qr}+\frac{p}{r}-\frac{3p}{q}\right)\geq0$$ or
$$\sum_{cyc}(2p^3+p^2q-3p^2r)\geq0$$ or
$$\sum_{cyc}(2p^3+p^2q-3q^2p)\geq0$$ or
$$\sum_{cyc}\left(p(p-q)(2p+3q)-\frac{5}{3}(p^3-q^3)\right)\geq0$$ or
$$\sum_{cyc}(p-q)^2(p+5q)\geq0.$$
Done!
A bit of easier way.
Let $x=\frac{b}{a}$ and $y=\frac{c}{b},$ where $a,$ $b$ and $c$ are positives.
Thus, by C-S we obtain:
$$\sum_{cyc}\frac{x}{(x+1)(y+2)}=\sum_{cyc}\frac{\frac{b}{a}}{\left(\frac{b}{a}+1\right)\left(\frac{c}{b}+2\right)}=$$
$$=\sum_{cyc}\frac{b^2}{(a+b)(c+2b)}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(a+b)(c+2b)}=\frac{1}{2}.$$
