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Algebra Difficulty 5.6 AIME, harder Prove it

Example 4 If aia_{i} have the same sign, a=∑ai≠0,n⩾2a=\sum a_{i} \neq 0, n \geqslant 2, n∈Nn \in \mathbf{N}. Then
∑ai2a−ai⩾n2n−1. \sum \frac{a_{i}}{2 a-a_{i}} \geqslant \frac{n}{2 n-1} .

Solutions — 2

Solution 1

 Prove: ∑ai2a−ai=∑ai−2a+2a2a−ai=−n+∑(2a−ai2a)−1=−n+∑(1−ai2a)−1⩾−n+n⋅(∑(1−ai/2a)n)−1=−n+n⋅nn−∑ai/2a=−n+n⋅nn−a/2a=n2n−1.\begin{array}{l}\text { Prove: } \sum \frac{a_{i}}{2 a-a_{i}}=\sum \frac{a_{i}-2 a+2 a}{2 a-a_{i}} \\ =-n+\sum\left(\frac{2 a-a_{i}}{2 a}\right)^{-1} \\ =-n+\sum\left(1-\frac{a_{i}}{2 a}\right)^{-1} \\ \geqslant-n+n \cdot\left(\frac{\sum\left(1-a_{i} / 2 a\right)}{n}\right)^{-1} \\ =-n+n \cdot \frac{n}{n-\sum a_{i} / 2 a} \\ =-n+n \cdot \frac{n}{n-a / 2 a} \\ =\frac{n}{2 n-1} .\end{array}

Solution 2

 Prove: ∑ai2a−ai=∑a1−2a+2a2a−ai=−n+∑(2a−ai2a)−1=−n+∑(1−ai2a)−1⩾−n+n⋅(∑(1−ai/2a)n)−1=−n+n⋅nn−∑ai/2a=−n+n⋅nn−a/2a=n2n−1.\begin{array}{l}\text { Prove: } \sum \frac{a_{i}}{2 a-a_{i}}=\sum \frac{a_{1}-2 a+2 a}{2 a-a_{i}} \\ =-n+\sum\left(\frac{2 a-a_{i}}{2 a}\right)^{-1} \\ =-n+\sum\left(1-\frac{a_{i}}{2 a}\right)^{-1} \\ \geqslant-n+n \cdot\left(\frac{\sum\left(1-a_{i} / 2 a\right)}{n}\right)^{-1} \\ =-n+n \cdot \frac{n}{n-\sum a_{i} / 2 a} \\ =-n+n \cdot \frac{n}{n-a / 2 a} \\ =\frac{n}{2 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.