# How to prove this inequality 4

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

• 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. Oct 22, 2015 at 10:25
• Keep in mind that $1-x_i=\sum_{i\neq j} x_j$ Oct 22, 2015 at 10:42
• 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 Oct 23, 2015 at 0:18

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

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

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