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 March 13th, 2019, 05:54 AM #1 Newbie   Joined: Mar 2019 From: USA Posts: 4 Thanks: 0 How can I prove this inequality? Let $a_1,a_2,...,a_n$ positive real numbers, where $n$ is an integer greater or equal to 3. Let $a_1+...+a_n=S_1$ and $a_1^2+...+a_n^2=S_2$. Prove that $$\sum_{i=1}^{n}\displaystyle \frac{S_1-a_i}{S_2-a_i^2}\geq n \cdot \displaystyle \frac{S_1}{S_2}$$ My trial. The inequality is in fact Chebyshev’s inequality: $$\left( \sum_{i=1}^{n}\displaystyle \frac{S_1-a_i}{S_2-a_i^2}\right)\cdot\left( \sum_{i=1}^{n}\left(S_2-a_i^2\right)\right) \geq n \cdot \left( \sum_{i=1}^{n}\left(S_1-a_i\right)\right)$$ The only problem is that it is impossible to prove that the sequences are inverse ordered.. so another method is required. If the inequality is true, please help me with a proof. If the inequality is not true, please give an example. Tags inequality, prove Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post StillAlive Calculus 5 September 2nd, 2016 11:45 PM walter r Real Analysis 2 January 15th, 2014 02:24 AM Albert.Teng Algebra 2 July 15th, 2013 04:06 AM Albert.Teng Algebra 4 July 13th, 2012 11:13 AM Albert.Teng Algebra 7 April 17th, 2012 04:53 AM

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