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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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. $\displaystyle 2|xy|\le x^2+y^2$ $\displaystyle |x+y+z|^2\le (|x|+|y|+|z|)^2=x^2+y^2+z^2+2|xy|+2|xz|+2|yz|$
Last edited by Plato; Oct 16th 2012 at 07:11 AM.
Ok, I am heading in the right direction... $\displaystyle 1+2x^2+2y^2+2z^2 \leq \sqrt{3}$
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 $\displaystyle |x+y+z|^2\le 3$
Ok, Maybe I am missing something, but what happened to 1?
Last edited by brucewayne; Oct 16th 2012 at 08:24 AM.
Originally Posted by brucewayne Ok, Maybe I am missing something, but what happens to 1? Come man do basic mathamatics. $\displaystyle 1+2x^2+2y^2+2z^2=1+2(x^2+y^2+z^2)=1+2(1)=3$
Dang it, I really do over analyze things at times. My sincerest apologies; I am trying to get better at this.
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