<s>[INST] Prove that for all $y,z\in\mathbb{R}^+$ it is true that $(y + z) (1 - 8 y z + z^2 + y^2 (1 + 9 z^2))\ge0$ Prove that for all $y,z\in\mathbb{R}^+$ it is true that
$$(y + z) (1 - 8 y z + z^2 + y^2 (1 + 9 z^2))\ge0$$

It is obvious that $y+z>0$. Then I tried to reduce $1 - 8 y z + z^2 + y^2 (1 + 9 z^2)$ to sum of squares. It seems to be hard, so I tried to expand it and I got 
$$y + y^3 + z - 7 y^2 z - 7 y z^2 + 9 y^3 z^2 + z^3 + 9 y^2 z^3\ge0$$
but what next? Is it possible to reduce it to sum of squares?
 [/INST] $$\begin{align}
&1-8yz+z^2+y^2\left(1+9z^2\right)\\
=\;&1-8yz+9(yz)^2+z^2+y^2\\
=\;&(3yz-1)^2-2yz+z^2+y^2\\
=\;&(3yz-1)^2+(z-y)^2\ge0
\end{align}$$
since it is the sum of squares
$y+z>0$ so their product
$$(y+z)\left(1-8yz+z^2+y^2\left(1+9z^2\right)\right)\ge0$$
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