# Prove the following sum inequality

If $x_1,x_2,\ldots,x_n$ are positive real numbers such that $\displaystyle \sum_{i = 1}^n x_i = 1$, prove that $$\displaystyle \sum_{i = 1}^n \dfrac{x_i}{\sqrt{1-x_i}} \geq \dfrac{\displaystyle \sum_{i = 1}^n \sqrt{x_i}}{\sqrt{n-1}}.$$

Seeing the $\displaystyle \sum_{i = 1}^n \sqrt{x_i}$ makes me think of Cauchy-Schwarz. But it might get bad as we have a square root. Therefore using a substitution might work. We can say $y_i = x_1+x_2+\cdots+x_n$ and then $1-x_i = y_i-x_i$. I am not sure how to make this substitution work.

First, recall that by Jensen's inequality $$\sqrt {\frac{x_1+...+x_n}{n} } \ge \frac{\sqrt x_1+...+\sqrt x_n } { n }$$ that using $x_1+...+x_n=1$ reduced to $$\sqrt n\ge \sqrt x_1+...+\sqrt x_n \tag{1}.$$
To reach the desired inequality we can use Jensen with $f(x)=\frac{x}{\sqrt {1-x}}$:
$$\frac{1}{n}\sum_i\frac{x_i}{\sqrt {1-x_i} } \ge \frac{\frac{\sum_i x_i}{n}}{\sqrt {1-\frac{\sum_i x_i}{n}}}=\frac{\frac{1}{n}}{\sqrt {1-\frac{1}{n}}}$$ then multiplying by $n$ on both sides $$\sum_i\frac{x_i}{\sqrt {1-x_i} } \ge \frac{\sqrt n}{\sqrt{n-1}}.$$ At this point we use $(1)$ and the inequality follows.
• Since $f(x)$ is convex, Jensen says exactly that $$f(\frac{\sum x_i}{n})\le \frac{f(\sum x_i)}{n}$$ – mrprottolo Jan 5 '16 at 2:05
• That is Jensen using $f(x)=\sqrt x$, that is concave. – mrprottolo Jan 5 '16 at 2:06