Inequality $\frac{1}{64}(a^{15} + b^{15} + c^{15}+ d^{15})^4 \ge \sum_{\rm {cyc}} (a b^2 c^4 d^8 )^4 $ Let real $a,b,c,d > 0$. Show that 
\begin{align}
\frac{1}{64}(a^{15} + b^{15} + c^{15}+ d^{15})^4 &\ge \sum_{\rm {cyc}}      (a b^2 c^4 d^8 )^4 \\
&=  (a b^2 c^4 d^8 )^4 +(b c^2 d^4 a^8 )^4 +(c d^2 a^4 b^8 )^4 +(d a^2 b^4 c^8 )^4 
\end{align}
This is obviously homogeneous. Equality appears at $a=b=c=d$, I haven't found other equality points. Applying AM-GM inequality to the LHS is too crude because it doesn't take into account the rising exponentials $(1,2,4,8)$ on the RHS. Indeed, AM-GM  leaves to prove 
\begin{align}
4 &\ge \sum_{\rm {cyc}}      \frac{c\cdot      d^{17}}{b^7 \cdot a^{11}}
\end{align}
which can be made to fail easily, e.g. by setting $a=b=c =0.01 \cdot d$.

Let $a b^2 c^4 d^8=t,$ $b c^2 d^4 a^8=z,$ $c d^2 a^4 b^8=y$ and $d a^2 b^4 c^8=x$.
Thus, $$\sum_{cyc}\frac{x^2}{y}=\sum_{cyc}\frac{d^2a^4b^8c^{16}}{cd^2a^4b^8}=\sum_{cyc}c^{15}$$ and we need to prove that
$$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{t}+\frac{t^2}{x}\geq\sqrt[4]{64(x^4+y^4+z^4+t^4)},$$ which is a known problem.
For example, see here: How to prove $\frac{1}{4}(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a})\ge \sqrt[4]{\frac{a^4+b^4+c^4+d^4}{4}}$
