Prove that $(x−2y+z)^2 \geq 4xz−8y$

inequalitymultivariable-calculuspolynomialsproof-verificationsum-of-squares-method

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.

Best Answer

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 $t\in[0,1]$.

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