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

This question already has an answer here:

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.

marked as duplicate by Martin Sleziak, Harish Chandra Rajpoot, user91500, user228113, SchrodingersCatFeb 7 '16 at 10:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• A minor point: The Hint should be $\frac{a+b}{2} \geq \sqrt{ab}$ – r9m Jul 21 '15 at 0:05
• Oh yeah woops, my mistake, thanks for pointing that out! – Jean Jul 21 '15 at 0:06
• 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 0:09
• the variables need to be all non negative – Dr. Sonnhard Graubner Jul 21 '15 at 3:22
• See also this question and other posts linked there. – Martin Sleziak Feb 7 '16 at 7:27

6 Answers

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

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

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 0:47

Notice

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

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

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

Implies

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

\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.

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.