Prove that $xy+yz+zx \leq x^2+y^2+z^2$

Question

Prove that $xy+yz+zx \leq x^2+y^2+z^2$ . Hint: Use $\frac{a+b}{2}\geq\sqrt{ab}$

First I tried using the hint by setting $a=x$ and $b=y+z$, however this results in the inequality: $$x^2+y^2+z^2 \geq 2xy-2yz+2zx $$ which isn't quite the same thing.

Then I tried starting with what we're trying to prove (to hopefully end up with a true statement), but then I get to this: $$(x+y+z)^2 \geq 3(xy+yz+zx)$$ and then I can't see what to do next.

This question is supposed to be straight forward, which is why I'm thinking there might be something wrong with it. Or I'm dumb.

As you can see I tagged this with Proof-strategy, so please don't bother writing down a full proof, just a few observations or hints are enough.

A minor point: The Hint should be $\frac{a+b}{2} \geq \sqrt{ab}$ — r9m, Jul 21 '15 at 00:05
Set $a = x^2$, $b= y^2$ first. You get $x^2+y^2 \ge 2xy$ (by the Hint). Do so for other $y,z$ and $z,x$ and add the three inequalities up :) — r9m, Jul 21 '15 at 00:09

score 10 · Accepted Answer · answered Jul 21 '15 at 00:08

10

$$ \frac{x^2 + y^2}{2} \geq xy $$ $$ \frac{y^2 + z^2}{2} \geq yz $$ $$ \frac{z^2 + x^2}{2} \geq zx $$ Now add these 3 inequalities side by side.

answered Jul 21 '15 at 00:08

Prism

11,782

score 5 · Answer 2 · answered Jul 21 '15 at 00:08

5

Hint: Use $(x-y)^2+(y-z)^2+(z-x)^2\ge 0$.

answered Jul 21 '15 at 00:08

André Nicolas

514,336

score 4 · Answer 3 · answered Jul 21 '15 at 00:43

An alternative using complex numbers.

Let $$ \phi = -\tfrac{1}{2} + \tfrac{1}{2} \mathbf{i} \sqrt{3}. $$ Write $$ \psi = x + \phi y + \phi^* z. $$

As we have $$ \psi \psi* \ge 0, $$ we get $$ \big( x + \phi y + \phi^* z \big) \big( x + \phi y + \phi^* z \big)^* \ge 0. $$

Whence $$ x^2 + \big(\phi^*\phi\big) y^2 + \big(\phi^*\phi\big) z^2 + \big(\phi^* + \phi \big) x y + \big(\phi^* + \phi \big) y z + \big(\phi^* + \phi \big) z x. $$

But $$ \phi^*\phi = 1, \hspace{2em} \phi^* + \phi = -1. $$

So we obtain $$ x^2 + y^2 + z^2 - xy - yz - zx \ge 0. $$

Oh wow, very creative! Also 'whence'! I'll definitely try your approach the next time I'm solving something similar. — Jean, Jul 21 '15 at 00:47

score 3 · Answer 4 · answered Jul 21 '15 at 00:09

3

Notice

$$ x^2 + y^2 \geq 2xy $$

$$ y^2 + z^2 \geq 2yz $$

$$ z^2 + x^2 \geq 2 zx $$

Implies

$$ 2x^2 + 2 y^2 + 2 z^2 \geq 2xy + 2yz + 2zx $$

answered Jul 21 '15 at 00:09

score 2 · Answer 5 · answered Jul 21 '15 at 00:11

2

\begin{eqnarray*} \frac{x^2+y^2}{2} \geq xy \\ \frac{z^2+y^2}{2} \geq yz \\ \frac{x^2+z^2}{2} \geq xz \end{eqnarray*} Adding them up we get the desired result.

answered Jul 21 '15 at 00:11

Discretizer

100

score 1 · Answer 6 · answered Jul 21 '15 at 00:12

It's a simple problem but here is one way to do it following hint: $$ xy\leq|xy|=\sqrt{x^2y^2}\leq\frac{x^2+y^2}{2}\implies xy\leq\frac{x^2+y^2}{2}. $$ The inequality $\sqrt{x^2y^2}\leq\frac{x^2+y^2}{2}$ is where you use the hint. Now repeat these steps 2 more time: one for the pair $\{y,z\}$ and another for $\{z,x\}$. It remains then to add the three results.

Prove that $xy+yz+zx \leq x^2+y^2+z^2$

6 Answers6