Let $x,y,z$ be nonnegative real numbers such that $x+z\leq2$
Prove that, and determine when equality holds.
$(x−2y+z)^2 \geq 4xz−8y$
Please correct me if my methods are incorrect or would lead nowhere.
I tried expanding the LHS of the inequality getting
$x^2+4y^2+z^2-4xy-4yz+2xz \geq 4xz-8y$
And got lost as to how I should manipulate the inequality to find something true through rough work.
After I tried manipulating
$x+z\leq2$ subtract 2
$x+z-2\leq0$ since $y\ge 0$
$x+z-2\le y$ subtract 2y and add 2 to both sides
And again lost sight of how I could manipulate the inequalities.