How do I show that $\frac a{1 - a^2} + \frac b{1 - b^2} + \frac c{1 - c^2} \ge \frac {3 \sqrt 3}2$

hint: Lets use $a = \tan\left(\frac{A}{2}\right)$. You can define $b, c$ similarly, then your inequality becomes: $\tan(A) + \tan(B) + \tan(C) \geq 3\sqrt{3}$, with $0 < A,B,C < \dfrac{\pi}{2}$ and $A+B+C = \pi$. And this inequality is standard result of convexity of $\tan(x)$ over $\left(0,\frac{\pi}{2}\right)$.

It's obviously wrong! Try $a=1.01$ and $b=c=\sqrt{2.0201}-1.01$.