## Wednesday, January 24, 2007

### Log Rolling. Topic: Inequalities. Level: AIME.

Problem: Given reals $a, b, c \ge 2$, show that $\log_{a+b}(c)+\log_{b+c}(a)+\log_{c+a}(b) \ge \frac{3}{2}$.

Solution: Recall the change-of-base identity for logs. We can rewrite the LHS as

$\frac{\log{c}}{\log{(a+b)}}+\frac{\log{a}}{\log{(b+c)}}+\frac{\log{b}}{\log{(c+a)}}$.

Note, however that $(a-1)(b-1) \ge 1 \Rightarrow ab \ge a+b \Rightarrow \log{a}+\log{b} = \log{(ab)} \ge \log{(a+b)}$, so it remains to show that

$\frac{\log{c}}{\log{a}+\log{b}}+\frac{\log{a}}{\log{b}+\log{c}}+\frac{\log{b}}{\log{c}+\log{a}} \ge \frac{3}{2}$,

which is true by Nesbitt's Inequality. QED.

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Comment: A good exercise in using log identities and properties to achieve a relatively simple result. Once we made the change-of-base substitution, seeing the $\frac{3}{2}$ should clue you in to Nesbitt's. That led to the inequality $\log{a}+\log{b} \ge \log{(a+b)}$, which was easily proven given the conditions of the problem.

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Practice Problem: Given reals $a, b, c \ge 2$, find the best constant $k$ such that

$\log_{a+b+c}(a)+\log_{a+b+c}(b)+\log_{a+b+c}(c) \ge k$.