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Algebra Difficulty 6.3 National olympiad Prove it

Example 11.12 Let a1,a2,⋯ ,ana_{1}, a_{2}, \cdots, a_{n} be positive real numbers, satisfying
a1+a2+⋯+an=1a1+1a2+⋯+1ana_{1}+a_{2}+\cdots+a_{n}=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}

Prove that
1n−1+a1+1n−1+a2+⋯+1n−1+an⩾1\frac{1}{n-1+a_{1}}+\frac{1}{n-1+a_{2}}+\cdots+\frac{1}{n-1+a_{n}} \geqslant 1

Solution

Let bi=1n−1+ai,i=1,2,⋯ ,nb_{i}=\frac{1}{n-1+a_{i}}, i=1,2, \cdots, n, then bin−1bib_{i}\frac{n-1}{b_{i}}

Thus,
∑j×i1−(n−1)bi1−(n−1)bj>(n−1)1−(n−1)bibi\sum_{j \times i} \frac{1-(n-1) b_{i}}{1-(n-1) b_{j}}>(n-1) \frac{1-(n-1) b_{i}}{b_{i}}

Summing the above equation for i=1,2,⋯ ,ni=1,2, \cdots, n,
∑i=1n∑j≠i1−(n−1)bi1−(n−1)bj>(n−1)∑1−(n−1)bibi\sum_{i=1}^{n} \sum_{j \neq i} \frac{1-(n-1) b_{i}}{1-(n-1) b_{j}}>(n-1) \sum \frac{1-(n-1) b_{i}}{b_{i}}

That is,
∑j=1n∑j×n1−(n−1)bi1−(n−1)bj>(n−1)∑1−(n−1)bibi\sum_{j=1}^{n} \sum_{j \times n} \frac{1-(n-1) b_{i}}{1-(n-1) b_{j}}>(n-1) \sum \frac{1-(n-1) b_{i}}{b_{i}}

By the assumption,
∑j≠i(1−(n−1)bi)(n−1)∑i=1n1−(n−1)bibi\sum_{j \neq i}\left(1-(n-1) b_{i}\right)(n-1) \sum_{i=1}^{n} \frac{1-(n-1) b_{i}}{b_{i}}

Contradiction! Hence, the original proposition is proved!

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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.