Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Possible Duplicate:
Cauchy-Schwarz inequality and three-letter identities (exercise 1.4 from “The Cauchy-Schwarz Master Class”)

Is it true for all $x, y, z > 0$ that

$$ x + y + z \leq 2 \left\{ \frac{x^2}{y+z} + \frac{y^2}{x+z} + \frac{z^2}{x+y} \right\} $$

This is an exercise (1.4) in "The Cauchy-Schwarz Master Class: An Introduction to the Art of mathematical Inequalities"

The solution suggests applying C-S to $$ x + y + z = \frac{x}{\sqrt{y+z}}\sqrt{y+z} + \frac{y}{\sqrt{x+z}}\sqrt{x+z} + \frac{z}{\sqrt{x+y}}\sqrt{x+y} $$

share|cite|improve this question

marked as duplicate by Martin Sleziak, Simon S, Rudy the Reindeer, Asaf Karagila, Henning Makholm Feb 3 '12 at 14:05

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.

up vote 2 down vote accepted

It is correct $ x + y + z = \frac{x}{\sqrt{y+z}}\sqrt{y+z} + \frac{y}{\sqrt{x+z}}\sqrt{x+z} + \frac{z}{\sqrt{x+y}}\sqrt{x+y}\le \left(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{y+x}\right)^{1/2}(2x+2y+2z)^{1/2}.$ Square both sides, and cancel out $x+y+z$.

share|cite|improve this answer
How embarrassing. Yes. – Simon S Feb 2 '12 at 14:25
No embarrassing. I make silly mistakes often... – Sunni Feb 2 '12 at 22:13

Not the answer you're looking for? Browse other questions tagged or ask your own question.