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Let $x_1,x_2,...,x_n$ be positive real numbers ; $n\geq 2$, such that $\displaystyle\sum_{i=1}^n x_i=1$.

How to prove that $\displaystyle\sum_{i=1}^n \sqrt{\dfrac{1-x_i}{x_i}}\geq (n-1)\displaystyle\sum_{i=1}^n \sqrt{\dfrac{x_i}{1-x_i}}$ ?

Thank in advances.

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  • $\begingroup$ Have you try something? You can try to change xi step by step to make it look like your expression and look what happen to 1. $\endgroup$
    – Deliss
    Oct 22, 2015 at 10:25
  • $\begingroup$ Keep in mind that $1-x_i=\sum_{i\neq j} x_j$ $\endgroup$ Oct 22, 2015 at 10:42
  • $\begingroup$ Let $\frac{1-x_i}{x_i}=a_i$, so that $\sum\frac{1}{1+a_i}=1$ and we need $\sum \sqrt{a_i}\ge (n-1)\sum\sqrt{\frac{1}{a_i}}$. This is then an old inequality, see for example page three here: vjimc.osu.cz/hist/j12solutions.pdf $\endgroup$
    – Apple
    Oct 23, 2015 at 0:18

1 Answer 1

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Suppose that $$x_1 \ge x_2 \ge \cdots \ge x_{n - 1} \ge x_1 \ge 0$$

$$\iff [\cdots] \iff \frac{1}{\sqrt{x_1(1 - x_1)}} \le \frac{1}{\sqrt{x_2(1 - x_2)}} \le \cdots \le \frac{1}{\sqrt{x_{n - 1}(1 - x_{n - 1})}} \le \frac{1}{\sqrt{x_n(1 - x_n)}}$$

Using the Chebyshev inequality, we have that

$$\left(\sum^n_{i = 1}x_i\right)\left[\sum_{cyc}\frac{1}{\sqrt{x_i(1 - x_i)}}\right] \ge n\left[\sum^n_{i = 1}x_1 \cdot \frac{1}{\sqrt{x_i(1 - x_i)}}\right]$$

$$\sum^n_{i = 1}\sqrt{\frac{1 - x_i}{x_i}} + \sum^n_{i = 1}\sqrt{\frac{x_i}{1 - x_i}} \ge n\sum^n_{i = 1}\sqrt{\frac{x_i}{1 - x_i}} \iff \sum^n_{i = 1}\sqrt{\frac{1 - x_i}{x_i}} \ge (n - 1)\left(\sum^n_{i = 1}\sqrt{\frac{a_i}{1 - a_i}}\right)$$

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