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March 13th, 2019, 05:54 AM   #1
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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
$$\sum_{i=1}^{n}\displaystyle \frac{S_1-a_i}{S_2-a_i^2}\geq n \cdot \displaystyle \frac{S_1}{S_2}$$

<b>My trial.</b> The inequality is in fact <i>Chebyshev’s inequality</i>:

$$\left( \sum_{i=1}^{n}\displaystyle \frac{S_1-a_i}{S_2-a_i^2}\right)\cdot\left( \sum_{i=1}^{n}\left(S_2-a_i^2\right)\right) \geq n \cdot \left( \sum_{i=1}^{n}\left(S_1-a_i\right)\right)$$

<i>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.</i>
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