prove inequality $2(x+z)^3<(3x+z)(x+3z)$ Let $0<x<1,0<z<1$. Then
$$
2(x+z)^{3}<(3x+z)(x+3z)
$$
This checks out numerically, but I don't know why. 

In the proof I'll use the fact that : $x^3<x^2$, $z^3<z^2$ ,$x^2<x$, $z^2<z$ and $2xz<x^2+z^2$. We have
\begin{align}
2(z+x)^3=2x^3+2z^3+6x^2z+6z^2x&< 2x^2+2z^2+12xz\\ &=2x^2+2z^2+10xz+2xz\\
&<2x^2+2z^2+10xz+z^2+x^2\\
&=(3x+z)(x+3z)
\end{align}
