Maths Olympiad Prep

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Algebra Difficulty 6.0 National olympiad Prove it

Similar to 8, if a1,a2,…,an∈R+,∑i=1nai=1a_{1}, a_{2}, \ldots, a_{n} \in R_{+}, \sum_{i=1}^{n} a_{i}=1, then:
∑i=1n(ai1−ai+∑i=1nai)≥n2n−1\sum_{i=1}^{n}\left(\frac{a_{i}}{1-a_{i}+\sum_{i=1}^{n} a_{i}}\right) \geq \frac{n}{2 n-1}

Solution

Proof: First, the left side of the original inequality needs to be adjusted to ∑i=1n[ai2(∑i=1nai)−ai]\sum_{i=1}^{n}\left[\frac{a_{i}}{2\left(\sum_{i=1}^{n} a_{i}\right)-a_{i}}\right] or ∑i=1n(21−ai+∑i=1nai)−n\sum_{i=1}^{n}\left(\frac{2}{1-a_{i}+\sum_{i=1}^{n} a_{i}}\right)-n, then apply the Cauchy-Schwarz inequality. □\square

If we add 1 to both the numerator and denominator of each term, the inequality not only has a minimum value but also a maximum value.

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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.