Tricky geometry proof If a,b,c belong to the interval $(0,1)$ and $ab + ca + bc = 1$, prove that 
$$\frac{a}{1-a^2} + \frac{b}{1-b^2} + \frac{c}{1-c^2}\ge\frac{3^{3/2}}{2}$$
How would you go about solving such a problem?

Another way: $(a+b+c)^2 \ge 3(ab+bc+ca) =3 \implies s=\dfrac{a+b+c}3 \ge \dfrac1{\sqrt3}$.
Now use $t \mapsto \dfrac{t}{1-t^2}$ is convex and Jensen's inequality to get
$$LHS \ge 3\frac{s}{1-s^2} \ge \frac{3\sqrt3}2=RHS$$
