Inequality

• Oct 16th 2012, 05:20 AM
brucewayne
Inequality
Let $\displaystyle x,y,z \in R$ such that $\displaystyle x^2+y^2+z^2=1$. Prove $\displaystyle |x+y+z| \leq \sqrt{3}$. I thinking along the lines of $\displaystyle AGM \leq GM$.
• Oct 16th 2012, 07:00 AM
Plato
Re: Inequality
Quote:

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

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

$\displaystyle |x+y+z|^2\le (|x|+|y|+|z|)^2=x^2+y^2+z^2+2|xy|+2|xz|+2|yz|$
• Oct 16th 2012, 07:39 AM
brucewayne
Re: Inequality
Ok, I am heading in the right direction...

$\displaystyle 1+2x^2+2y^2+2z^2 \leq \sqrt{3}$
• Oct 16th 2012, 07:43 AM
Plato
Re: Inequality
Quote:

Originally Posted by brucewayne
Ok, I am heading in the right direction...

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

Factor out the 2 in the last three terms.
You have
$\displaystyle |x+y+z|^2\le 3$
• Oct 16th 2012, 07:59 AM
brucewayne
Re: Inequality
Ok, Maybe I am missing something, but what happened to 1?
• Oct 16th 2012, 08:05 AM
Plato
Re: Inequality
Quote:

Originally Posted by brucewayne
Ok, Maybe I am missing something, but what happens to 1?

Come man do basic mathamatics.
$\displaystyle 1+2x^2+2y^2+2z^2=1+2(x^2+y^2+z^2)=1+2(1)=3$
• Oct 16th 2012, 08:09 AM
brucewayne
Re: Inequality
Dang it, I really do over analyze things at times. My sincerest apologies; I am trying to get better at this.