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Algebra Difficulty 7.3 National olympiad, round 2 Prove it

Example 9 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}^{*}, we have (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)

Solution

(a+b)n−an−bn+1=(ab)n−an−bn+1=(an−1)(bn−1)=(a−1)(b−1)(an−1+an−2+⋯+a+1)(bn−1+bn−2+⋯+b+1)=(an−1+an−2+⋯+a+1)(bn−1+bn−2+⋯+b+1)⩾[(ab)n−12+(ab)n−22+⋯+(ab)12+1]2⩾[4n−12+4n−22+⋯+412+1]2=(2n−1+2n−2+⋯+2+1)2=(2n−1)2\begin{array}{l} (a+b)^{n}-a^{n}-b^{n}+1=(a b)^{n}-a^{n}-b^{n}+1=\left(a^{n}-1\right)\left(b^{n}-1\right)= \\ (a-1)(b-1)\left(a^{n-1}+a^{n-2}+\cdots+a+1\right)\left(b^{n-1}+b^{n-2}+\cdots+b+1\right)= \\ \left(a^{n-1}+a^{n-2}+\cdots+a+1\right)\left(b^{n-1}+b^{n-2}+\cdots+b+1\right) \geqslant \\ {\left[(a b)^{\frac{n-1}{2}}+(a b)^{\frac{n-2}{2}}+\cdots+(a b)^{\frac{1}{2}}+1\right]^{2} \geqslant} \\ {\left[4^{\frac{n-1}{2}}+4^{\frac{n-2}{2}}+\cdots+4^{\frac{1}{2}}+1\right]^{2}=} \\ \left(2^{n-1}+2^{n-2}+\cdots+2+1\right)^{2}=\left(2^{n}-1\right)^{2} \end{array}

That is,
(a+b)n−an−bn⩾22n−2n+1(a+b)^{n}-a^{n}-b^{n} \geqslant 2^{2 n}-2^{n+1}

Prove that because 1a+1b=1\frac{1}{a}+\frac{1}{b}=1, so a+b=aba+b=a b, (a−1)(b−1)=1(a-1)(b-1)=1, and because 1a+\frac{1}{a}+ 1b=1⩾21ab\frac{1}{b}=1 \geqslant 2 \sqrt{\frac{1}{a b}}, so ab⩾4a b \geqslant 4. Therefore,
(a+b)n−an−bn+1=(ab)n−an−bn+1=(an−1)(bn−1)=(a−1)(b−1)(an−1+an−2+⋯+a+1)(bn−1+bn−2+⋯+b+1)=(an−1+an−2+⋯+a+1)(bn−1+bn−2+⋯+b+1)⩾[(ab)n−12+(ab)n−22+⋯+(ab)12+1]2⩾[4n−12+4n−22+⋯+412+1]2=(2n−1+2n−2+⋯+2+1)2=(2n−1)2\begin{array}{l} (a+b)^{n}-a^{n}-b^{n}+1=(a b)^{n}-a^{n}-b^{n}+1=\left(a^{n}-1\right)\left(b^{n}-1\right)= \\ (a-1)(b-1)\left(a^{n-1}+a^{n-2}+\cdots+a+1\right)\left(b^{n-1}+b^{n-2}+\cdots+b+1\right)= \\ \left(a^{n-1}+a^{n-2}+\cdots+a+1\right)\left(b^{n-1}+b^{n-2}+\cdots+b+1\right) \geqslant \\ {\left[(a b)^{\frac{n-1}{2}}+(a b)^{\frac{n-2}{2}}+\cdots+(a b)^{\frac{1}{2}}+1\right]^{2} \geqslant} \\ {\left[4^{\frac{n-1}{2}}+4^{\frac{n-2}{2}}+\cdots+4^{\frac{1}{2}}+1\right]^{2}=} \\ \left(2^{n-1}+2^{n-2}+\cdots+2+1\right)^{2}=\left(2^{n}-1\right)^{2} \end{array}

That is,
(a+b)n−an−bn⩾22n−2n+1(a+b)^{n}-a^{n}-b^{n} \geqslant 2^{2 n}-2^{n+1}

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