## Tuesday, December 20, 2005

### Back Again? Topic: Inequalities. Level: Olympiad.

Problem: (2003 USAMO - #5) Let $a,b,c$ be positive real numbers. Prove that

$\frac{(2a+b+c)^2}{2a^2+(b+c)^2}+\frac{(2b+c+a)^2}{2b^2+(c+a)^2}+\frac{(2c+a+b)^2}{2c^2+(a+b)^2} \le 8$.

Solution: Well first of all we notice that the inequality is homogeneous. Therefore, we can set $a+b+c$ to an arbitrary value. Note that using $a+b+c = 3$ works particularly well.

Then the inequality becomes

$\displaystyle \sum \frac{(3+a)^2}{2a^2+(3-a)^2} = \sum \frac{a^2+6a+9}{3a^2-6a+9} = \sum \left(\frac{1}{3}+\frac{8a+6}{3(a-1)^2+6}\right)$.

Well, noticing that $\frac{8a+6}{3(a-1)^2+6} \le \frac{8a+6}{6} = 1+\frac{4a}{3}$ we have

$\displaystyle \sum \left(\frac{1}{3}+\frac{8a+6}{3(a-1)^2+6}\right) \le \sum \frac{4}{3}(1+a) = \frac{4}{3}(3+a+b+c) = \frac{4}{3}(6) = 8$

as desired. QED.

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Comment: This was a pretty cool problem, too. The most important part was splitting the fraction and getting rid of the quadratic term. After both the squared terms are gone, it's pretty simple to finish off with only linear and constant terms.