AGM Inequality Proof 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 :)
 A: This is the solution to the old version of the problem:
The right hand side is $4xy-8y=4y(x-2)$ and hence non-positive, since $y$ is non-negative $x$ is at most $2$. But the left hand side is a square and hence non-negative. So the inequality follows. 

For the revised version:
Let $u=x+z$, so the LHS is $(u-2y)^2$. Notice that swapping value between $x$ and $z$ does not affect $u$, but by the AGM the quantity $xz$ is at most $(\frac u2)^2$. So the RHS is at most $u^2-8y$. 
So the problem reduces to showing $(u-2y)^2\geq u^2-8y$. This is easy if $y=0$, so we may assume $y\neq 0$. Expand the left, cancel $u^2$ and divide by $4y$ to reduce to $-u+y\geq -2$, or $y\geq u-2$, which is true. So we're done.
A: This way doesn't quite use the AM-GM inequality but is pretty similar: First note that $x$ and $z$ are symmetrical in this problem, so without loss of generality you can assume that $z \leq x$. Since $x + z \leq 2$ this means that $z \leq 1$.
Note that $(a + b)^2 \geq 4ab$ for all $a$ and $b$. (This is almost the same as the AM-GM inequality). In view of your left hand side this is a natural form to use. The question is what to use as $a$ and what to use as $b$. Looking at what you are trying to show, I use $a = x - 2y$ and $b = z$. I obtain
$$(x - 2y + z)^2 \geq 4xz - 8yz$$
Since $z \leq 1$, we have $-8yz \geq -8y$. Substituting this in we get the needed
$$(x - 2y + z)^2 \geq 4xz - 8y$$
As to how you'd know to do this.. I got the second to last equation first, then tried to see how to convert this into the form you want. It was clear I needed $z \leq 1$ so then I used the condition $x + z \leq 2$ as above.
A: Use $\rm\ (x-2y+z)^2 = 4xz-8y + 4y(2-x-z) + (x-z)^2 + 4y^2\:.\:$ But last 3 summands are $\ge 0\:.\ $ Note this proof requires only that $\rm\ \ y(2-x-z)\ge 0\:.$
A: EDIT: This solves the earlier version of the question.
Here's a more "brute-force" approach which seems silly given JDH's solution, but might help you attack similar questions.
First of all, we see that $z$ is just there to confuse us. Suppose $x+z$ is held constant - that determines the LHS. In order to prove the inequality, we should consider the maximal RHS. Since the RHS is monotone increasing with $x$, we might as well put $z=0$.
Now expand the square and simplify $LHS-RHS \geq 0$ to get $$x^2 + 4y^2 - 8xy + 8y \geq 0.$$ What do we do with this unwieldy expression? Suppose there were some "bad" values of $y$, for which some $x \in [0,2]$ would be negative. Since everything is continuous, we could find the "critical" values of $y$ by trying both endpoints. Putting $x = 0$ we get the critical value $y = 0$ (as a solution to the equation $LHS - RHS = 0$). Putting $x = 2$ we get the critical value $y = 1$: $$4 + 4y^2 - 16y + 8y = 4y^2 - 8y + 4 = 4(y-1)^2.$$ This suggests substituting $y = Y + 1$: $$x^2 + 4(Y+1)^2 - 8x(Y+1) + 8(Y+1) = x^2 - 8xY - 8x + 4Y^2 + 16Y + 12.$$ That doesn't look any better, does it? We know that $4Y^2$ should be isolated. We quickly notice the subexpression $-8xY+16Y = 8(2-x)Y \geq 0$. What remains is the factorable $x^2 - 8x + 12 = (x-6)(x-2) \geq 0$, and in total we get $$(x-6)(x-2) + 8(2-x)Y + 4Y^2 \geq 0.$$
A: we have to prove
$$(x - 2y + z)^2 \geq 4xz - 8y$$
expanding L.H.S
$$(x+z)^{2}+ 4y^{2} - 4y(x+z) $$
now,
$(x+z)^2 \geq 4xz$ ..............{A.M G.M .in.equality}
$4y(x+z) \geq 8y$ ,..........................since $x+z \leq 2$
hence,
$$(x+z)^{2}+ 4y^{2} - 4y(x+z) \geq 4xz -8y +4y^{2} \geq 4xy -8y$$
