2010-2011 Mock USAJMO Problems/Solutions/Problem 2

Problem 2

Let $x, y, z$ be positive real numbers such that $x+y+z = 1$ . Prove that $\frac{3x+1}{y+z}+\frac{3y+1}{z+x}+\frac{3z+1}{x+y}\ge\frac{4}{2x+y+z}+\frac{4}{x+2y+z}+\frac{4}{x+y+2z},$ with equality if and only if $x=y=z$ .

Solution 1

First, we change the terms using the relationship $x+y+z=1$ :

$\sum_{cyc}\frac{3x+1}{y+z} - \sum_{cyc}\frac{4}{2x+y+z}$ $= \sum_{cyc}(\frac{4}{y+z}-1) - \sum_{cyc}\frac{4}{1+x}= 4\sum_{cyc} (\frac{1}{1-x} - \frac{1}{1+x}) - 9$ $= 8\sum_{cyc} \frac{x}{1-x^2} - 9$

Then, by Cauchy-Schwarz Inequality, one has: $\sum_{cyc}\frac{x}{1-x^2} \cdot \sum_{cyc} (x-x^3) \ge (x+y+z)^2 = 1$

And by AM-GM, one has: $\sum_{cyc} (x-x^3) = 1- \sum_{cyc} x^3 \le 1- 3(\frac{x+y+z}{3})^3 = \frac{8}{9}$

Therefore, $8\sum_{cyc} \frac{x}{1-x^2} - 9 \ge 8 \cdot \frac{1}{\sum_{cyc} (x-x^3)} - 9 \ge 8 \cdot \frac{9}{8} - 9 = 0,$

where both equations are true if and only if $x=y=z=\frac{1}{3}$ .

edited by User:lightest 17:35, 24 April 2012 (EDT)

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2010-2011 Mock USAJMO Problems/Solutions/Problem 2

Problem 2

Solution 1