Inequality

**brucewayne** · October 16th 2012, 05:20 AM

Let $x,y,z \in R$ such that $x^2+y^2+z^2=1$ . Prove $|x+y+z| \leq \sqrt{3}$ . I thinking along the lines of $AGM \leq GM$ .

**Plato** · October 16th 2012, 07:00 AM

Originally Posted by brucewayne

Let $x,y,z \in R$ such that $x^2+y^2+z^2=1$ .
Prove $|x+y+z| \leq \sqrt{3}$ .

Here are the facts you need.
$2|xy|\le x^2+y^2$

$|x+y+z|^2\le (|x|+|y|+|z|)^2=x^2+y^2+z^2+2|xy|+2|xz|+2|yz|$

**brucewayne** · October 16th 2012, 07:39 AM

Ok, I am heading in the right direction...

$1+2x^2+2y^2+2z^2 \leq \sqrt{3}$

**Plato** · October 16th 2012, 07:43 AM

Originally Posted by brucewayne

Ok, I am heading in the right direction...

$1+2x^2+2y^2+2z^2 \leq \sqrt{3}$

Factor out the 2 in the last three terms.
You have
$|x+y+z|^2\le 3$

**brucewayne** · October 16th 2012, 07:59 AM

Ok, Maybe I am missing something, but what happened to 1?

**Plato** · October 16th 2012, 08:05 AM

Originally Posted by brucewayne

Ok, Maybe I am missing something, but what happens to 1?

Come man do basic mathamatics.
$1+2x^2+2y^2+2z^2=1+2(x^2+y^2+z^2)=1+2(1)=3$

**brucewayne** · October 16th 2012, 08:09 AM

Dang it, I really do over analyze things at times. My sincerest apologies; I am trying to get better at this.

Thread: Inequality