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Algebra Difficulty 6.3 National olympiad Prove it

24. Beckenbach (E. F. Beckenbach) and Bellman (R. Bellman) recorded an interesting conclusion by Fan Rui in a journal: If 0<xi⩽12,i=1,2,⋯ ,n0<x_{i} \leqslant \frac{1}{2}, i=1,2, \cdots, n, then
∏i=1nxi∏i=1n(1−xi)⩽(∑i=1nxi)n(∑i=1n(1−xi))n\frac{\prod_{i=1}^{n} x_{i}}{\prod_{i=1}^{n}\left(1-x_{i}\right)} \leqslant \frac{\left(\sum_{i=1}^{n} x_{i}\right)^{n}}{\left(\sum_{i=1}^{n}\left(1-x_{i}\right)\right)^{n}}

Equality holds if and only if all xix_{i} are equal.

Solution

24. We can prove it by backward induction. If x1≠x2x_{1} \neq x_{2}, we have
x1x2(1−x1)(1−x2)−(x1+x2(1−x1)+(1−x2))2=(x1−x2)2(x1+x2−1)(1−x1)(1−x2)((1−x1)+(1−x2))2x1x2⋯xn−1A(1−x1)(1−x2)⋯(1−xn−1)(1−Aˉ)\begin{aligned} & \frac{x_{1} x_{2}}{\left(1-x_{1}\right)\left(1-x_{2}\right)}-\left(\frac{x_{1}+x_{2}}{\left(1-x_{1}\right)+\left(1-x_{2}\right)}\right)^{2} \\ = & \frac{\left(x_{1}-x_{2}\right)^{2}\left(x_{1}+x_{2}-1\right)}{\left(1-x_{1}\right)\left(1-x_{2}\right)\left(\left(1-x_{1}\right)+\left(1-x_{2}\right)\right)^{2}} & \frac{x_{1} x_{2} \cdots x_{n-1} A}{\left(1-x_{1}\right)\left(1-x_{2}\right) \cdots\left(1-x_{n-1}\right)(1-\bar{A})} \end{aligned}

Therefore, it also holds for n−1n-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.