Maths Olympiad Prep

Track / Stage 6 / 8 of 400 #1008 of 1964

Problem 1008

National olympiad, first round
Algebra Difficulty 6.0 Prove it

Example 6 Prove: (1+13)(1+15)⋯(1+12n−1)>2n+12\left(1+\frac{1}{3}\right)\left(1+\frac{1}{5}\right) \cdots\left(1+\frac{1}{2 n-1}\right)>\frac{\sqrt{2 n+1}}{2}. Here, n∈Nn \in N, and n⩾2n \geqslant 2.

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.

Official solution

Prove: Construct the sequence {Tn}:1+13,(1+13)\left\{T_{n}\right\}: 1+\frac{1}{3},\left(1+\frac{1}{3}\right)
- (1+15),⋯ ,(1+13)(1+15)⋯(1+\left(1+\frac{1}{5}\right), \cdots,\left(1+\frac{1}{3}\right)\left(1+\frac{1}{5}\right) \cdots(1+ 12n+1),⋯\left.\frac{1}{2 n+1}\right), \cdots, then
TnTn−1=1+12n+1=(2n+2)2(2n+1)2>(2n+3)(2n+1)(2n+1)2=2n+32n+1. Hence ,Tn2n+3>Tn−12n+1. \begin{array}{c} \frac{T_{n}}{T_{n-1}}=1+\frac{1}{2 n+1}=\sqrt{\frac{(2 n+2)^{2}}{(2 n+1)^{2}}} \\ >\sqrt{\frac{(2 n+3)(2 n+1)}{(2 n+1)^{2}}}=\sqrt{\frac{2 n+3}{2 n+1}} . \\ \text { Hence }, \frac{T_{n}}{\sqrt{2 n+3}}>\frac{T_{n-1}}{\sqrt{2 n+1}} . \end{array}

Therefore, the sequence {Tn−12n+1}\left\{\frac{T_{n-1}}{\sqrt{2 n+1}}\right\} is monotonically increasing, and its first term
 is T15=435. Thus Tn−1⩾4352n+1>2n−12, so (1+13)(1+15)⋯(1+12n−1)>2n+12. \begin{array}{l} \text { is } \frac{T_{1}}{\sqrt{5}}=\frac{4}{3 \sqrt{5}} . \\ \text { Thus } T_{n-1} \geqslant \frac{4}{3 \sqrt{5}} \sqrt{2 n+1}>\frac{\sqrt{2 n-1}}{2}, \\ \text { so }\left(1+\frac{1}{3}\right)\left(1+\frac{1}{5}\right) \cdots\left(1+\frac{1}{2 n-1}\right) \\ \quad>\frac{\sqrt{2 n+1}}{2} . \end{array}

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.