How do I show that $\frac a{1 - a^2} + \frac b{1 - b^2} + \frac c{1 - c^2} \ge \frac {3 \sqrt 3}2$ For $0 \lt a, b, c \lt 1$, if $ab + bc + ca = 1$, show that $$\frac a{1 - a^2} + \frac b{1 - b^2} + \frac c{1 - c^2} \ge \frac {3 \sqrt 3}2.$$
I want to use trigonometric substitution:
For the angles $A, B, C$ of any acute triangle, $$\tan A + \tan B + \tan C = \tan A \tan B \tan C,$$ $$\frac 1{\tan A \tan B} + \frac 1{\tan B \tan C} + \frac 1{\tan C \tan A} = 1.$$
Also, $\tan A, \tan B, \tan C \gt 0$. So I substitute $a, b, c$ for $\frac 1{\tan A}, \frac 1{\tan B}, \frac 1{\tan C}$ respectively. Then the inequality in question becomes $$\frac {\tan A}{1 - \tan^2 A} + \frac {\tan B}{1 - \tan^2 B} + \frac {\tan C}{1 - \tan^2 C} \le -\frac {3 \sqrt 3}2.$$
Here $A, B, C \not = \frac {\pi}4$ since $a, b, c \not = 1$.
By the trigonometric identity $\tan 2A = \frac {2 \tan A}{1 - \tan^2 A}$, we have
$$\tan 2A + \tan 2B + \tan 2C \le -3 \sqrt 3,$$
where $0 \lt A, B, C \lt \frac {\pi}2$, $A, B, C \not = \frac {\pi}4$, and $A + B + C = \pi$.
How do I proceed?
Edit: The restriction $a, b, c \lt 1$ was added after the question had received some answers, thanks to Michael Rozenberg, who pointed out this mistake.
 A: 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 $\left(0,\frac{\pi}{2}\right)$.
A: It's obviously wrong! Try $a=1.01$ and $b=c=\sqrt{2.0201}-1.01$.
A: We can prove $S=\sum\tan A=\prod\tan A$
Using AM GM inequality if $\tan A,\tan B,\tan C\ge0,$ $$\dfrac{\sum\tan A}3\ge\sqrt[3]{\prod\tan A}$$
$$\iff\dfrac S3\ge\sqrt[3]S\implies\left(\dfrac S3\right)^3\ge S\iff S^2\ge27$$ as $S\ne0$
Can you take it from here?
A: If you only want to prove it with AM-GM, I will give one proof using AM-GM and Cauchy-Schwarz. 
Note that the given condition implies that $a+b+c\geq\sqrt 3$. In addition, I will use the following two inequalities. $$(a+b+c)(ab+bc+ca)\leq a^3+b^3+c^3+6abc$$ and $$abc\leq\big(\dfrac{ab+bc+ca}{3}\big)^{\frac{3}{2}}.$$ Now, 
$$\sum\limits_{a,b,c} \dfrac{a}{1-a^2}\sum\limits_{a,b,c} a(1-a^2)\geq (a+b+c)^2.$$ But $\sum\limits_{a,b,c} a(1-a^2) = (a+b+c)(ab+bc+ca)-a^3-b^3-c^3\leq abc\leq \dfrac{6}{3\sqrt 3}=\dfrac{2}{\sqrt 3}$. These two combined together implies that $$\sum\limits_{a,b,c} \dfrac{a}{1-a^2}\geq\dfrac{(a+b+c)^2}{\dfrac{2}{\sqrt 3}}\geq\dfrac{3 \sqrt 3}{2}.$$
 The first inequality is called Schur's inequality.  
EDIT. 
Actually an even easier solution is obtained by using $$\dfrac{a}{1-a^2}+\dfrac{9}{4}a(1-a^2)\geq 3a,$$ by AM-GM. Thus, $\sum\limits_{a,b,c} \dfrac{a}{1-a^2}\geq\dfrac{3}{4}\sum\limits_{a,b,c} a+\dfrac{9}{4}\sum\limits_{a,b,c} a^3\geq\dfrac{3}{4}\cdot\sqrt 3 + \dfrac{9}{4}\cdot\dfrac{1}{\sqrt 3} = \dfrac{3 \sqrt 3}{2}$. That $a+b+c\geq\sqrt 3$ and $a^3+b^3+c^3\geq\dfrac{1}{\sqrt 3}$ are trivial from the given condition.
