Prove $\sum ^n_{i=1} \frac{x_i}{\sqrt{1-x_i}}\geq \frac{1}{\sqrt{n-1}}\sum ^n_{i=1} x_i$

Question

If $\space x_1+x_2+\cdots+x_n=1$ and all $x_1,x_2,\cdots,x_n$ are positive and real numbers, prove:$$\sum ^n_{i=1} \frac{x_i}{\sqrt{1-x_i}}\geq \frac{1}{\sqrt{n-1}}\sum ^n_{i=1} x_i$$ Additional:we are allowed to use Cauchy(better to use it more than other inequalities),
AM-GM and other simple inequalities. However if you think the problem could not solved with allowed inequalities,use anything you want to solve question.

Things I have tried so far:

I can re-write inequality as:

$$\sum ^n_{i=1} \frac{x_i}{\sqrt{1-x_i}}\geq \frac{1}{\sqrt{n-1}}$$

Using Cauchy inequality:

$$\left(\sum ^n_{i=1} \frac{x_i}{\sqrt{1-x_i}}\right) \left(\sum ^n_{i=1} x_i\sqrt{1-x_i} \right)\geq \left(\sum ^n_{i=1} x_i\right)^2$$

Now my problem simplifies to proving this:

$$\frac{1}{\left(\sum\limits_{i=1}^n x_i\sqrt{1-x_i}\right)}\geq \frac{1}{\sqrt{n-1}}$$

$$\sqrt{n-1}\geq \left(\sum ^n_{i=1} x_i\sqrt{1-x_i}\right)$$

And I stuck here.

Answer 1

As discussed on the comments, as the terms only make sense for $n \ge 2$, we can just compare the two sides term by term.

$1 - x_i \le 1 \le n - 1$, so $\frac{x_i}{\sqrt{1-x_i}} \ge \frac{x_i}{\sqrt{n-1}}$

Now just sum each side for i going from 1 to n.