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A have found a very interesting inequality in a Romanian magazine which I use to prepare for the Lithuanian Mathematical Olympiad. Let $a_1,a_2,...,a_n$ be positive real numbers such that $$\frac {1} {1+a_1+a_2+...+a_{n-1}} + \frac {1} {1+a_1+a_2+...+a_{n-2}+a_n} + ... +\frac {1} {1+a_2+a_3+...+a_{n}} \ge 1.$$ Show that $$(n-1)(a_1+a_2+...+a_n) \ge 2 \sum_{1 \le i \lt j \le n} a_ia_j.$$ I have tried Jensen's inequality, but I didn't get to any result. In my proof, all I need to prove is that $a_1+a_2+...+a_n<n$.

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Denote $s=a_1+a_2+\ldots +a_n$ and $p=2 \sum_{1 \le i \lt j \le n} a_ia_j$. We need to show that $(n-1)s \ge p$ given $$\sum_{i=1}^n \frac{1}{1+s-a_i} \ge 1.$$ Rewriting the above inequality as $$\sum_{i=1}^n \left( \frac{1}{1+s-a_i} - 1 \right) \ge 1-n$$ or equivalently $$\sum_{i=1}^n \frac{s-a_i}{1+s-a_i} \le n-1.$$ Applying Cauchy-Schwarz inequality to the LHS of the last inequality: \begin{align} \sum_{i=1}^n \frac{s-a_i}{1+s-a_i} &= \sum_{i=1}^n \frac{(s-a_i)^2}{(1+s-a_i)(s-a_i)} \\ &\ge \frac{\left(\sum_{i=1}^n(s-a_i)\right)^2}{\sum_{i=1}^n(1+s-a_i)(s-a_i)} \\ &= \frac{(n-1)^2s^2}{(n-1)s + \sum_{i=1}^n (s-a_i)^2} \end{align} Therefore we have $$n-1 \ge \frac{(n-1)^2s^2}{(n-1)s + \sum_{i=1}^n (s-a_i)^2}$$ or equivalently $$(n-1)s + \sum_{i=1}^n (s-a_i)^2 \ge (n-1)s^2 \quad (*)$$ By noticing that $\sum_{i=1}^n a_i^2 = s^2 - p$, it can be easily seen that $\sum_{i=1}^n (s-a_i)^2 = (n-1)s^2-p$. Plugging this into $(*)$ we get $(n-1)s \ge p$, QED.

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  • $\begingroup$ Excellent proof, thank you so much! $\endgroup$ – I. Stefan May 3 '16 at 17:54

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