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 $a=b=c =0.01 \cdot d$.

The  answer to the corresponding problem (see Michael's comment) has been provided by me at math.stackexchange.com/a/3566860/317854 
Hence the problem is solved.
