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 $x=1/\sqrt[4]{3}$
 A: We have
\begin{align*}
 \frac{1}{1 - x^8}
 &= \frac{1}{8/9 + (1/9 - x^8)}\\
 &= \frac{9}{8}\cdot \frac{1}{1+ (9/8)(1/9 - x^8)}\\
 &\ge \frac98 \left(1 - \frac{9}{8}(1/9 - x^8)\right) \tag{1}\\
 &= \frac{9}{64}(7 + 9x^8)\\
 &\ge \frac{9}{64}\cdot 8\sqrt[8]{1^7 \cdot 9x^8} \tag{2}\\
 &= \frac98 \sqrt[8]{9} \, x
\end{align*}
where we have used $\frac{1}{1 + u} \ge 1 - u$ for all $-1 < u < 1$ in (1) and AM-GM in (2).
Thus, we have
$$\sum_{\mathrm{cyc}} \frac{x^3}{1 - x^8} \ge \frac98 \sqrt[8]{9}\, (x^4 + y^4 + z^4) = \frac98 \sqrt[8]{9}.$$
Also, when $x = y = z = 1/\sqrt[4]{3}$, we have $x^4 + y^4 + z^4 = 1$ and $\sum_{\mathrm{cyc}} \frac{x^3}{1 - x^8} = \frac98 \sqrt[8]{9}$.
Thus, the minimum of $\sum_{\mathrm{cyc}} \frac{x^3}{1 - x^8}$ is $\frac98 \sqrt[8]{9}$.
A: 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.
