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.

  • $\begingroup$ Is it a typo that your sums start at $i=0?$ $\endgroup$ Jul 29, 2014 at 12:08
  • $\begingroup$ yea ,what an error.i will fix it now. $\endgroup$ Jul 29, 2014 at 12:09
  • $\begingroup$ Am I missing something? For $n \ge 2$ we can just compare the two sides term by term, $1-x_i \le 1 \le n-1$ so $\frac{x_i}{\sqrt{1-x_i}} \ge \frac{x_i}{\sqrt{n-1}}$. For n=1 both sides are infinite. $\endgroup$
    – Wonder
    Jul 29, 2014 at 12:16
  • $\begingroup$ @Wonder, n is always larger than 1.hmm your solution seems good.let me think about it. $\endgroup$ Jul 29, 2014 at 12:20
  • 1
    $\begingroup$ I think his problem was mainly that the indices were going from 0 to n here but 1 to n in the constraint. So intended meaning should be getting represented fine here. I have posted it as an answer like you suggested. $\endgroup$
    – Wonder
    Jul 29, 2014 at 12:28

1 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.

  • $\begingroup$ denominator should not be zero,So $1 - x_i < 1 \le n - 1$ $\endgroup$ Jul 29, 2014 at 12:37
  • $\begingroup$ Yes, so in fact this is always a strict inequality! $\endgroup$
    – Wonder
    Jul 29, 2014 at 12:38
  • $\begingroup$ So $\sum ^n_{i=1} \frac{x_i}{\sqrt{1-x_i}}= \frac{1}{\sqrt{n-1}}\sum ^n_{i=1} x_i$ is not possible in any situation? the $\geq $ on question make me unsure somehow... $\endgroup$ Jul 29, 2014 at 12:41
  • $\begingroup$ It is easy to see that it is not possible. I think probably $\ge$ sign makes it easier to solve using methods where you multiply both sides with products of denominators like how you were trying it. $\endgroup$
    – Wonder
    Jul 29, 2014 at 12:44

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