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 🙂
Best Answer
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.