Prove that $(x−2y+z)^2 \geq 4xz−8y$

Question

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.

Answer 1

Note that $y\geq 0$ and $x+z\leq 2$ imply $$(x-2y+z)^2+8y\geq (x-2y+z)^2+4(x+z)y=x^2+4y^2+z^2+2xz\,.$$ Thus, $$(x-2y+z)^2+8y\geq (x-z)^2+4y^2+4xz\geq 4xz\,,$$ whence $$(x-2y+z)^2\geq 4xz-8y\,.$$ The equality holds if and only if $(x,y,z)=(t,0,t)$ for some $t\in[0,1]$.

Answer 2

Because $$(x-2y+z)^2+8y-4xz\geq(x-2y+z)^2+4y(x+z)-4xz=$$ $$=x^2+4y^2+z^2-2xz=(x-z)^2+4y^2\geq0.$$ The equality occurs for $y=0$,$x=z\geq0$ and $x+z\leq2$.

Answer 3

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.