Showing an inequality using Cauchy-Schwarz I managed to solve the following inequality using AM-GM:
$$
\frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)} \geq \frac{3}{4}
$$
provided that $a,b,c >0$ and $abc=1$.
However it was hinted to me that this could also be solved with Cauchy-Schwarz inequality but I have not been able to find a solution using it and I'm really out of ideas.

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:
$$\sum_{cyc}\frac{a}{(a+1)(b+1)}=\sum_{cyc}\frac{\frac{y}{x}}{\left(\frac{y}{x}+1\right)\left(\frac{z}{y}+1\right)}=\sum_{cyc}\frac{y^2}{(x+y)(y+z)}\geq\frac{(x+y+z)^2}{\sum\limits_{cyc}(x+y)(y+z)}.$$
Id est, it's enough to prove that:
$$\frac{(x+y+z)^2}{\sum\limits_{cyc}(x+y)(y+z)}\geq\frac{3}{4}$$ or
$$4(x+y+z)^2\geq3\sum_{cyc}(x+y)(y+z)$$ or 
$$4\sum_{cyc}(x^2+2xy)\geq3\sum_{cyc}(x^2+3xy)$$ or$$\sum_{cyc}(x^2-xy)\geq0$$ or $$\sum_{cyc}(x-y)^2\geq0.$$
