Given $x^4+y^4+z^4=1$ Find the minimum value of $\sum_{cyc}\frac{x^3}{1-x^8}$. Let $x,y,z$ be positive reals such that  $x^4+y^4+z^4=1$. Find the minimum value of
$$\sum_{cyc}\frac{x^3}{1-x^8}$$
First I tried Jensen. Let $$f(x)=\frac{x^3}{1-x^8}$$
Then $f$ is convex on $\mathbb R^+$, using Jensen $$\frac{1}{3}\sum_{cyc}f(x)\ge f\left(\frac{x+y+z}{3}\right)$$
but we don't have enough information about $x+y+z$.
By the way my guess for the minimum value is at $x=1/\sqrt[4]{3}$

Consider the function $$f(x)=\frac{1}{1-x^8}$$
And note that $f$ is convex on $[0,1[$ hence it lies above its tangent line, $$f(x)\ge f'(a)(x-a)+f(a)$$
set $a=1/\sqrt[4]{3}$
$$\frac{1}{1-x^8}\ge \frac{9}{8}\sqrt[4]{3}x$$
Meaning $$\frac{x^3}{1-x^8}\ge \frac{9}{8}\sqrt[4]{3}x^4$$
Now cycle sum to finish.
