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?
 A: Let that $x,y,z$ are the angles of a triangle, such that:
$$a=\tan(x/2),b=\tan(y/2),c=\tan(z/2),$$
You might be knowing these properties about a triangle:
$$\tan(x/2)\tan(y/2)+\tan(y/2)\tan(z/2)+\tan(z/2)\tan(x/2)=1$$
Which is given in question.Also:
$$x+y+z=\pi$$
By the way, maximum value of an angle evident from question is $\pi/2$, so that it is an acute triangle.Left Hand Side ( LHS ) of the inequality converts to this using the property
$$\frac12(\tan x+\tan y+\tan z),\quad\frac{2\tan x}{1-\tan^2x}=\tan2x$$
Now maybe you can minimize this.
You can also use this property of a triangle:
$$\tan x+\tan y+\tan z=\tan x\tan y\tan z$$
So LHS of the inequality becomes $$\frac12\tan x\tan y\tan z$$

Hint: 
By some symmetry reasons minima must occur at the point when all angle are equal,i.e. $x=y=z=\pi/3$.Putting values:
$$\frac12(\sqrt3+\sqrt3+\sqrt3)=\frac12(\sqrt3\sqrt3\sqrt3)=\frac32\sqrt3$$
A: 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
$$LHS \ge 3\frac{s}{1-s^2} \ge \frac{3\sqrt3}2=RHS$$
