Tasty problem to be digested

Relevant wiki: Jensen's Inequality

Here's my approach. First, we set $x^4=a, y^4=b, z^4=c\Rightarrow a+b+c=1$ , then the expression is $\sum_{cyc}\frac{\sqrt[4]{a^3}}{1-a^2}$ Now consider this function $f(t)=\frac{\sqrt[4]{t^3}}{1-t^2}$ $f'(t)=\frac{5t^2+3}{4\sqrt[4]{t}(t^2-1)^2}>0; \forall x\in\mathbb{R}$ So $f(t)$ increase in the interval $(0;1)$ , making it a convex function. Now by Jensen's inequality , we have $f(a)+f(b)+f(c)\geq 3f\bigg(\frac{a+b+c}{3}\bigg)=3f\big(\frac{1}{3}\big)$ $\therefore\sum_{cyc} \frac {x^3}{1 - x^8}\geq\frac{3\sqrt[4]{(\frac{1}{3})^3}}{1-(\frac{1}{3})^2}=\frac{9\sqrt[4]{3}}{8}\approx 1.480$ The equality holds when $x=y=z=\sqrt[4]{\frac{1}{3}}$