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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 $$ x + y + z = \frac{x}{\sqrt{y+z}}\sqrt{y+z} + \frac{y}{\sqrt{x+z}}\sqrt{x+z} + \frac{z}{\sqrt{x+y}}\sqrt{x+y} $$

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

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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 $x+y+z$.

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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

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