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. 


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.
