Olympiad Maths Prep

Track / Stage 7 / 180 of 300 #1580 of 2000

Problem 1580

National olympiad second round; IMO P1/P4
Algebra Difficulty 7.4 Prove it

Example 8 Given that a,ba, b are positive real numbers, and 1a+1b=1\frac{1}{a}+\frac{1}{b}=1, prove: for every n∈N∗n \in \mathbf{N}^{*}, (a+b)n−an−bn⩾22n−2n+1(a+b)^{n}-a^{n}-b^{n} \geqslant 2^{2 n}-2^{n+1}. (1988 National High School Mathematics League Question)

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Prove that for a=sec⁡2θ,b=csc⁡2θ,0<θ<π2a=\sec ^{2} \theta, b=\csc ^{2} \theta, 0<\theta<\frac{\pi}{2}, we have 1a+1b=1\frac{1}{a}+\frac{1}{b}=1. The original inequality is equivalent to
(an−1)(bn−1)⩾(2n−1)2\left(a^{n}-1\right)\left(b^{n}-1\right) \geqslant\left(2^{n}-1\right)^{2}

which is
(sec⁡2nθ−1)(csc⁡2nθ−1)⩾(2n−1)2\left(\sec ^{2 n} \theta-1\right)\left(\csc ^{2 n} \theta-1\right) \geqslant\left(2^{n}-1\right)^{2}

or
[(tan⁡2θ+1)n−1]=[(cot⁡2θ+1)n−1]=⩾(2n−1)2\left[\left(\tan ^{2} \theta+1\right)^{n}-1\right]=\left[\left(\cot ^{2} \theta+1\right)^{n}-1\right]=\geqslant\left(2^{n}-1\right)^{2}

By the binomial theorem and the Cauchy-Schwarz inequality, we get:
[(tan⁡2θ+1)n−1]⋅[(cot⁡2θ+1)n−1]=(∑k=1nCnktan⁡2kθ)⋅(∑k=1nCnkcot⁡2kθ)⩾(∑k=1nCnktan⁡kθcot⁡kθ)2=(∑k=1nCnk)2=(∑k=0nCnk−1)2=(2n−1)2\begin{aligned} {\left[\left(\tan ^{2} \theta+1\right)^{n}-1\right] \cdot } & {\left[\left(\cot ^{2} \theta+1\right)^{n}-1\right]=\left(\sum_{k=1}^{n} C_{n}^{k} \tan ^{2 k} \theta\right) \cdot\left(\sum_{k=1}^{n} C_{n}^{k} \cot ^{2 k} \theta\right) \geqslant } \\ & \left(\sum_{k=1}^{n} C_{n}^{k} \tan ^{k} \theta \cot ^{k} \theta\right)^{2}=\left(\sum_{k=1}^{n} C_{n}^{k}\right)^{2}= \\ & \left(\sum_{k=0}^{n} C_{n}^{k}-1\right)^{2}=\left(2^{n}-1\right)^{2} \end{aligned}

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