Prove that $$\sum_{cyc}\frac{x}{1-x^2}\ge \frac{3\sqrt{3}}{2}$$ where $xy+yz+zx=1$ and $0< x,y,z <1$
I have a proof :Let $2A,2B,2C$ be the angles of an acute angled triangle, also let $x=\tan A,y=\tan B,z=\tan C$ easy to check $0< x,y,z <1$. $$\sum_{cyc}\frac{x}{1-x^2}=\frac{1}{2} \left(\sum_{cyc} \tan 2A \right)\ge \frac{3\sqrt{3}}{2}$$
Because in a triangle with angles $2A ,2B,2C$ $$\sum_{cyc} \tan 2A\ge 3\sqrt{3}$$
I am however looking for alternative proofs!
We have $$\frac{x}{1-x^2} - \left(3x-\frac{\sqrt{3}}{2}\right) = \frac{(2x+\sqrt 3)(3x-\sqrt 3)^2}{6(1-x^2)} \geqslant 0.$$ So $$\sum \frac{x}{1-x^2} \geqslant 3(x+y+z)-\frac{3\sqrt{3}}{2}\geqslant 3\sqrt{3(xy+yz+zx)} - \frac{3\sqrt{3}}{2}$$ $$=\frac{3\sqrt{3}}{2}.$$ Equality occur when $x=y=z=\frac{1}{\sqrt 3}.$
$$f(x)=\frac{x}{1-x^2} \implies f''(x)=\frac{2x}{(x^2+3)}{(1-x^2)^3}>0, 0<x<1.$$ So by Jensen's Inequality if folloes that $$\frac{f(x)+f(y)+f(z)}{3}\ge f\left(\frac{x+y+z}{3}\right), 0<x,y,z<1.~~~~(1)$$ Also $$xy+yz+zx=1 \implies (x+y+z)^2 \ge 3~ \implies p \ge \sqrt{3}~~~(2)$$ From (1), we can write $$F=\frac{x}{1-x^2}+\frac{y}{1-y^2}+\frac{z}{1-z^2} \ge 3 \frac{p/3}{1-p^2/9}$$ From (2) $$ p \ge \sqrt{3} \implies p^2/9 \ge 1/3 \implies 1-p^2/9 \le 2/3 \implies \frac{1}{1-p^2/9}\ge \frac{3}{2} \implies \frac{p}{1-p^2/9} \ge \frac{3\sqrt{3}}{2}.$$ Hence, $$F \ge \frac{3\sqrt{3}}{2}.$$
