<s>[INST] 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.
 [/INST] 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 $\left(0,\frac{\pi}{2}\right)$.
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