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. 
 A: $$
2(x+z)^3=2x^{3}+6x^{2}z+6xz^{2}+2z^{3}<2x^{3}+(5z+1)x^{2}+(5x+1)z^{2}+2z^{3}<
$$
$$
2x^{2}+(5z+1)x^{2}+(5x+1)z^{2}+2z^{2}=3x^{2}+5x^{2}z+5xz^{2}+3z^{2}<
$$
$$
3x^{2}+10xz+3z^{2}=(3x+z)(x+3z)
$$
A: 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}
A: $$(x+z)^3-(3x+z)(x+3z)<(x+z)^3-(x+z+2)^2$$
We know that $0<x+z<2$, so consider the polynomial $P(t)=t^3-(t+2)^2$ in the intervl $[0,2]$. Its derivative is $3t^2-2t-4=3(t-\frac13)^2-4.33\ldots$, that has only a root in the interval $[0,2]$. It is easy to see that this root is a minimum for $P$. Then the maxima of $P$ are in the ends of the interval.
$$P(0)=-4$$
$$P(2)=-8$$
This shows the inequality.
A: 
Expanding, our claim is that
  $$ 2x^3 + 6xz^2 + 6x^2z + 2z^3 < 3x^2 + 3z^2 + 10xz$$
  $0 < x < 1, 0 < z < 1$.
  Note that we can write the RHS as $$2x^2 + 2z^2 + x^2 + z^2 + 10xz $$

We see that 
$$ 2x^3 < 2x^2 \\ 2z^3 < 2z^2 \\ 6xz^2 < 5xz + z^2 \\ 6x^2z < 5xz + x^2$$
By adding these together our claim follows.
