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Algebra Difficulty 7.3 National olympiad, round 2 Prove it

18. (1) Given 0⩽x⩽1,n0 \leqslant x \leqslant 1, n is a positive integer, prove the inequality: (1+x)n⩾(1−x)n+(1+x)^{n} \geqslant(1-x)^{n}+ 2nx(1−x2)n−12 n x \sqrt{\left(1-x^{2}\right)^{n-1}}. (1988 Irish Mathematical Olympiad Problem)

Solution

 18. (1)(1+x)n−(1−x)n=[(1+x)−(1−x)]⋅[(1+x)n−1+(1+x)n−2(1−x)+⋯+(1+x)(1−x)n−2+(1−x)n−1]=2x[(1+x)n−1+(1+x)n−2(1−x)+⋯+(1+x)(1−x)n−2+(1−x)n−1]⩾2x⋅n⋅(1+x)n−1(1+x)n−2(1−x)⋅⋯⋅(1+x)(1−x)n−2(1−x)n−1n=2x⋅n⋅[(1+x)(1+x)](n−1)+(n−2)+⋯+2+1n=2nx(1−x2)n−1.\begin{array}{l}\text { 18. }(1)(1+x)^{n}-(1-x)^{n}=[(1+x)-(1-x)] \cdot\left[(1+x)^{n-1}+(1+\right. \\ \left.x)^{n-2}(1-x)+\cdots+(1+x)(1-x)^{n-2}+(1-x)^{n-1}\right]=2 x\left[(1+x)^{n-1}+(1+\right. \\ \left.x)^{n-2}(1-x)+\cdots+(1+x)(1-x)^{n-2}+(1-x)^{n-1}\right] \geqslant 2 x \cdot n \cdot \\ \sqrt[n]{(1+x)^{n-1}(1+x)^{n-2}(1-x) \cdot \cdots \cdot(1+x)(1-x)^{n-2}(1-x)^{n-1}}=2 x \cdot n \cdot \\ \sqrt[n]{[(1+x)(1+x)]^{(n-1)+(n-2)+\cdots+2+1}}=2 n x \sqrt{\left(1-x^{2}\right)^{n-1}} .\end{array}

Translating the text into English:

 18. (1)(1+x)n−(1−x)n=[(1+x)−(1−x)]⋅[(1+x)n−1+(1+x)n−2(1−x)+⋯+(1+x)(1−x)n−2+(1−x)n−1]=2x[(1+x)n−1+(1+x)n−2(1−x)+⋯+(1+x)(1−x)n−2+(1−x)n−1]⩾2x⋅n⋅(1+x)n−1(1+x)n−2(1−x)⋅⋯⋅(1+x)(1−x)n−2(1−x)n−1n=2x⋅n⋅[(1+x)(1+x)](n−1)+(n−2)+⋯+2+1n=2nx(1−x2)n−1.\begin{array}{l}\text { 18. }(1)(1+x)^{n}-(1-x)^{n}=[(1+x)-(1-x)] \cdot\left[(1+x)^{n-1}+(1+\right. \\ \left.x)^{n-2}(1-x)+\cdots+(1+x)(1-x)^{n-2}+(1-x)^{n-1}\right]=2 x\left[(1+x)^{n-1}+(1+\right. \\ \left.x)^{n-2}(1-x)+\cdots+(1+x)(1-x)^{n-2}+(1-x)^{n-1}\right] \geqslant 2 x \cdot n \cdot \\ \sqrt[n]{(1+x)^{n-1}(1+x)^{n-2}(1-x) \cdot \cdots \cdot(1+x)(1-x)^{n-2}(1-x)^{n-1}}=2 x \cdot n \cdot \\ \sqrt[n]{[(1+x)(1+x)]^{(n-1)+(n-2)+\cdots+2+1}}=2 n x \sqrt{\left(1-x^{2}\right)^{n-1}} .\end{array}

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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.