<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] Hint: Use completing the square for $-8yz+z^2+y^2$  which gives 
$$-8yz+z^2+y^2=y^2+z^2-2yz-6yz=(y-z)^2-6yz$$
Now, \begin{align}
1-8yz+z^2+y^2+9y^2z^2&= 1+(y-z)^2-6yz+9y^2z^2\\
&=(y-z)^2+(3yz)^2-2.3yz.1+1\\
&=(y-z)^2+(3yz-1)^2
\\
&\geq 0
\end{align}
</s>