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 $.
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.
ReplyDeletewe 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
ReplyDeleteLol nice. Just made the title up randomly.
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