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 $.

3 comments:

  1. The new problem is easier, I think. It becomes abc \ge (a + b + c)^k. At a = b = c = 2 we see we require k \le log_6 (8), and it's pretty clear that if we increase a, b, c, the LHS increases faster than the RHS.

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  2. we talked about logrolling in Health Science today, aka turning someone in bed with a draw sheet, used on people with spinal injuries or just out of spinal surgery. Thought it was a funny coincidence

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  3. Lol nice. Just made the title up randomly.

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