# 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 :)

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.

• Hey sorry i made a mistake, the right hand side is 4xz - 8y instead of 4xy - 8y. Dunno if you can take another look, appreciate it, thanks. – 1337holiday Jan 8 '11 at 2:52
• This is great!! I didnt know you can assume the maximum, for xz. How did you come to know that you can assume that? Because im new to this and would like to know what approach you use for these kinds of problems. Thanks so much, this really helped, simple and clear. – 1337holiday Jan 8 '11 at 3:38
• The idea is just that in order to show A is at least B, it suffices to show that A is at least as big as the biggest B could possibly be. In some sense, this is making the problem harder, since you are solving the worst possible case, but since you aim to solve all the cases anyway, you might as well put yourself in the worst possible case. And often, it happens in this worst case that simplifications occur and you can see why something is true. – JDH Jan 8 '11 at 3:42
• Great, i now have a better understanding of this and will keep this in mind when i attempt other problems. – 1337holiday Jan 8 '11 at 4:00

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.

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 $\rm\ \ y(2-x-z)\ge 0\:.$

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 $$(x-6)(x-2) + 8(2-x)Y + 4Y^2 \geq 0.$$

• Is it possible to do a simpler method for a beginner? I really appreciate the answer and if you can help me understand this question, i will grasp a better idea on how to approach these problems, thanks a lot. – 1337holiday Jan 8 '11 at 3:10

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, $$(x+z)^{2}+ 4y^{2} - 4y(x+z) \geq 4xz -8y +4y^{2} \geq 4xy -8y$$