Prove that $(x−2y+z)^2 \geq 4xz−8y$ 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
$x-2y+z\le 2-y$
And again lost sight of how I could manipulate the inequalities.

consider,
$(x-2y+z)^2-4xz+8y=x^2+4y^2+z^2-4xy-4yz+2xz-4xz+8y$
$=x^2+4y^2+z^2-4xy-4yz-2xz+8y=x^2+4y^2+z^2-4y(x+z)-2xz+8y$
$=(x-z)^2+4y^2+4y(2-(x+z))\geq 0$
because $y$ is non negative and $x+z\leq 2$.
Hence that inequality holds.
