I have been stuck on this one for hours.
Let $x$, $y$, $z$ be non-negative real numbers.
Also we know $x + z \leq 2$.
Prove the following:
$(x - 2y + z)^2 \geq 4xz - 8y$.
Apparently this can be proven with or without AGM, which is $xy \leq \left(\frac{x + y}{2}\right)^2$.
This is what I have done so far: \begin{align} ((x + z) - 2y)^2 &\geq 4xz - 8y\\ (x + z)^2 -4y(x + z) + 4y^2 &\geq 4xz - 8y &\quad&\text{expanded the squared term keeping }(x+z)\\ (x + z)^2 -4yx - 4yz + 4y^2 &\geq 4xz - 8y&&\text{now we have AGM}\\ \left(\frac{x + z}{2}\right)^2 -yx -yz + y^2 &\geq xz -2y \end{align} Rearranging we have $\left(\frac{x + z}{2}\right)^2 -yx -yz + 2y \geq xz - y^2.$
This is as far as I got and we also know that the told us this: $x + z \leq 2$.
Rearranging we get: $x + z - 2 \leq 0$; then multiply by $y$ to get: $yx + yz - 2y \leq 0$.
I am new to this proofs and if someone can guide me as to how to attempt these and whats the method to solve these question that would be really great, thanks :)