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Problem 1141

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Algebra Difficulty 5.0 Prove it Singapore Mathematical Olympiad (SMO) · Singapore

Let 0<a1<a2<⋯<an0 < a_1 < a_2 < \dots < a_n be real numbers. Prove that
(11+a1+11+a2+⋯+11+an)2≤1a1+1a2−a1+1a3−a2+⋯+1an−an−1. \left(\frac{1}{1+a_1} + \frac{1}{1+a_2} + \dots + \frac{1}{1+a_n}\right)^2 \le \frac{1}{a_1} + \frac{1}{a_2-a_1} + \frac{1}{a_3-a_2} + \dots + \frac{1}{a_n-a_{n-1}}.

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

By Cauchy-Schwarz inequality,
LHS≤(1a1+1a2−a1+⋯+1an−an−1)(a1(1+a1)2+a2−a1(1+a2)2+⋯+an−an−1(1+an)2). \text{LHS} \le \left( \frac{1}{a_1} + \frac{1}{a_2 - a_1} + \dots + \frac{1}{a_n - a_{n-1}} \right) \left( \frac{a_1}{(1+a_1)^2} + \frac{a_2 - a_1}{(1+a_2)^2} + \dots + \frac{a_n - a_{n-1}}{(1+a_n)^2} \right).
Note that a1(1+a1)2≤a11+a1\frac{a_1}{(1+a_1)^2} \le \frac{a_1}{1+a_1} and for i=2,…,ni=2, \dots, n,
ai−ai−1(1+ai)2≤ai−ai−1(1+ai−1)(1+ai)=11+ai−1−11+ai. \frac{a_i - a_{i-1}}{(1+a_i)^2} \le \frac{a_i - a_{i-1}}{(1+a_{i-1})(1+a_i)} = \frac{1}{1+a_{i-1}} - \frac{1}{1+a_i}.
Thus, after telescoping,
a1(1+a1)2+a2−a1(1+a2)2+⋯+an−an−1(1+an)2≤a11+a1+11+a1−11+an<1 \frac{a_1}{(1+a_1)^2} + \frac{a_2 - a_1}{(1+a_2)^2} + \dots + \frac{a_n - a_{n-1}}{(1+a_n)^2} \le \frac{a_1}{1+a_1} + \frac{1}{1+a_1} - \frac{1}{1+a_n} < 1
and we are done.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.