# Non totally symmetric inequality

How can I prove that, for all positive $a,b,c$ such that $abc=1$ we have $$S=\left(\frac{a}{1+ab}\right)^2+\left(\frac{b}{1+bc}\right)^2+\left(\frac{c}{1+ac}\right)^2\ge\frac{3}{4}?$$

By Jensen, I know $S\ge\frac{1}{3}\left(\frac{a}{1+ab}+\frac{b}{1+bc}+\frac{c}{1+ac}\right)^2$, so if i show that $\frac{a}{1+ab}+\frac{b}{1+bc}+\frac{c}{1+ac}\ge\frac{3}{2}$, I'll winish. Indeed, wolframalpha said me that this is true, but I don't know how to prove it.

Let $a=\frac{y}{x}$ and $b=\frac{z}{y}$, where $x$, $y$ and $z$ are positives.

Thus, $c=\frac{x}{z}$ and by C-S we obtain: $$\sum_{cyc}\frac{a}{1+ab}=\sum_{cyc}\frac{\frac{y}{x}}{1+\frac{y}{x}\cdot\frac{z}{y}}=\sum_{cyc}\frac{y}{x+z}=$$ $$=\sum_{cyc}\frac{y^2}{xy+yz}\geq\frac{(x+y+z)^2}{2(xy+xz+yz)}\geq\frac{3}{2},$$ where the last inequality it's just $$\sum_{cyc}(x-y)^2\geq0.$$ Done!