Maths Olympiad Prep

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Problem 896

AMC 12 late, AIME early
Combinatorics Difficulty 4.7 Prove it All-Soviet-Union Mathematical Olympiad · Soviet Union

nn numbers are written on a blackboard. Someone then repeatedly erases two numbers and writes half their arithmetic mean instead, until only a single number remains. If all the original numbers were 11, show that the final number is not less than 1/n1/n.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

Solution:

Put c=(a+b)/4c = (a + b)/4. We have 1/c=4/(a+b)1/a+1/b1/c = 4/(a + b) \leq 1/a + 1/b, so each move does not increase the sum of the reciprocals of the numbers. If the final number is kk, then the final sum of reciprocals is 1/k1/k. The initial sum is nn, so 1/kn1/k \leq n, or k1/nk \geq 1/n.

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