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 $\displaystyle AGM \leq GM$.

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- Oct 16th 2012, 05:20 AMbrucewayneInequality
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 AMPlatoRe: Inequality
- Oct 16th 2012, 07:39 AMbrucewayneRe: 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 AMPlatoRe: Inequality
- Oct 16th 2012, 07:59 AMbrucewayneRe: Inequality
Ok, Maybe I am missing something, but what happened to 1?

- Oct 16th 2012, 08:05 AMPlatoRe: Inequality
- Oct 16th 2012, 08:09 AMbrucewayneRe: Inequality
Dang it, I really do over analyze things at times. My sincerest apologies; I am trying to get better at this.