But it's Jensen.

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=61144&hilit=China+1989

Does this work? -- It's pretty cool because we don't really prove much of anything and we are able to conlcude-it feels a bit indirect Let $S=\displaystyle\sum_{i=1}^n{\frac{x_i}{\sqrt{1-x_i}}}$ Note by Rearrangement that, because $\{x_i\}_{1\le i\le n}$ and $\left\{\frac{1}{\sqrt{1-x_i}}\right\}_{1\le i\le n}$ are similarly arranged, \[ S\ge \frac{x_2}{\sqrt{1-x_1}}+\frac{x_3}{\sqrt{1-x_2}}+\cdot\cdot\cdot+\frac{x_1}{\sqrt{1-x_n}}\\ S\ge \frac{x_3}{\sqrt{1-x_1}}+\frac{x_4}{\sqrt{1-x_2}}+\cdot\cdot\cdot+\frac{x_2}{\sqrt{1-x_n}}\\ .\\ .\\ .\\ S\ge \frac{x_n}{\sqrt{1-x_1}}+\frac{x_1}{\sqrt{1-x_2}}+\cdot\cdot\cdot+\frac{x_{n-1}}{\sqrt{1-x_n}} \] Summing and dividing by $(n-1)$, we have $S\ge\frac{1}{n-1}(\sqrt{1-a_1}+\sqrt{1-a_2}+\cdot\cdot\cdot+\sqrt{1-a_n})$. Hence, the given inequality is true if we can show that $\frac{1}{n-1}(\sqrt{1-a_1}+\sqrt{1-a_2}+\cdot\cdot\cdot+\sqrt{1-a_n})\ge\frac{\sum_{i=1}^{n}\sqrt{x_i}}{\sqrt{n-1}}\;\;\;(*)$ Now, note by Cauchy Schwarz that \[ (\sqrt{1-a_1}+\sqrt{1-a_2}+\cdot\cdot\cdot+\sqrt{1-a_n})\left(\sum^{n}_{i=1}\frac{x_i}{\sqrt{1-x_i}}\right)\ge(\sqrt{a_1}+\cdot\cdot\cdot+\sqrt{a_n})^2 \] which, by dividing by the sum of the square roots on the LHS, implies \[ \sum^{n}_{i=1}\frac{x_i}{\sqrt{1-x_i}}\ge\left(\frac{\sqrt{a_1}+\cdot\cdot\cdot+\sqrt{a_n}}{\sqrt{1-a_1}+\sqrt{1-a_2}+\cdot\cdot\cdot+\sqrt{1-a_n}}\right)(\sqrt{a_1}+\cdot\cdot\cdot+\sqrt{a_n}) \] Hence the given inequality is true if we can show that $\left(\frac{\sqrt{a_1}+\cdot\cdot\cdot+\sqrt{a_n}}{\sqrt{1-a_1}+\sqrt{1-a_2}+\cdot\cdot\cdot+\sqrt{1-a_n}}\right)\ge\frac{1}{\sqrt{n-1}}\;\;\;(**)$ Now, however, note that $(*)$ and $(**)$ are opposites and thus one of them must be true. Thus, the given inequality is true.

By Chebyshev's inequality on sorted sequences \[ a_i = x_i \] and \[b_i = \frac{1}{\sqrt{1-x_i}} \], we can see that \[\sum^{n}_{i=1}\dfrac{x_i}{\sqrt{1-x_i}}\ge \sum^{n}_{i=1}\frac{\sqrt{1-x_i}}{n} \] Summing up and subtracting the original first term, we can obtain $\frac{1}{n-1}(\sqrt{1-a_1}+\sqrt{1-a_2}+\cdot\cdot\cdot+\sqrt{1-a_n})\ge\frac{\sum_{i=1}^{n}\sqrt{x_i}}{\sqrt{n-1}}$ But this follows easily from summing up $\sqrt{1-a_1} = \sqrt{a_2 + a_3 \cdots + a_n} \ge \frac{\sqrt{a_2} + \sqrt{a_3} \cdots + \sqrt{a_n}}{\sqrt{n-1}}$ and similar, so we are done.

here we can see the LHS has a symmetry so we can try to break it in order to get inequalities in small terms and then we add up and get the result. we write this as \[\sum_{j=1}^n\frac{a_j}{\sqrt{a_j-1}}=\sum_{j=1}^n\frac{1}{\sqrt{1-a_j}}-\sum_{j=1}^n\sqrt{1-a_j}\]by AM-GM we can write this as \[\sum_{j=1}^n\frac{1}{\sqrt{1-a_j}}\geq n\left\{\frac{1}{\prod_{j=1}^n\sqrt{1-a_j}}\right\}^{1/n}\]\[n\sqrt{\frac{1}{\prod_{j=1}^n(1-a_j)^{1/n}}}\]here we have that \[\left(\prod_{j=1}^n(1-a_j)\right)^{1/n}\leq \frac{1}{n}\sum_{j=1}^n(1-a_j)=\frac{1}{n}(n-1)=\frac{n-1}{n} \]\[\implies \sum_{j=1}^n\frac{1}{\sqrt{1-a_j}}\geq n\sqrt{\frac{n}{n-1}}\]by cauchy's inequality we can write this as \[\sum_{j=1}^n\sqrt{1-a_j}\leq \left(\sum_{j=1}^nj\right)^{1/2}\left(\sum_{j=1}^n(1-a_j)\right)^{1/2}=\sqrt{n} \sqrt{\sum_{j=1}^n(1-a_j)}=\sqrt{n(n-1)}\]so now we can write \[\sum_{j=1}^n\frac{1}{\sqrt{1-a_j}}-\sum_{j=1}^n\sqrt{1-a_j}\geq n\sqrt{\frac{n}{n-1}}-\sqrt{n(n-1)}=\sqrt{\frac{n}{n-1}} \]by cauchy's inequality we can write this as \[\left(\sum_{j=1}^nj\right)\left(\sum_{j=1}^na_j\right)\geq \left(\sum_{j=1}^n\sqrt{a_j}\right)^2\]\[\implies \sqrt{n}\geq\sum_{j=1}^n\sqrt{a_j}\]now the orignal problem we can write that as \[\sum_{j=1}^n\frac{a_j}{\sqrt{a_j-1}}\geq \frac{\sqrt{n}}{\sqrt{n-1}}\geq \frac{1}{\sqrt{n-1}}\sum_{j=1}^n\sqrt{a_j}\]and we are done $\blacksquare$

Let $f(x) = \frac{x}{\sqrt{1-x}}$ . Then $f^{''}(x) = \frac{4-x}{4(1-x)^{\frac{5}{2}}} > 0 $ . Then $$f(x_{1})+...+f(x_{n}) \geq nf(\frac{x_{1}+...+x_{n}}{n}) = nf(\frac{1}{n}) = \frac{\sqrt{n}}{\sqrt{n-1}}$$Now we need to prove $ \frac{\sqrt{n}}{\sqrt{n-1}} \geq \dfrac{\sum_{i=1}^{n}\sqrt{x_i}}{\sqrt{n-1}}$ . ( ) ( ) is easy by $C-S$ . $\blacksquare$

jred wrote: Let $x_1, x_2, \dots ,x_n$ ($n\ge 2$) be positive real numbers satisfying $\sum^{n}_{i=1}x_i=1$. Prove that:\[\sum^{n}_{i=1}\dfrac{x_i}{\sqrt{1-x_i}}\ge \dfrac{\sum_{i=1}^{n}\sqrt{x_i}}{\sqrt{n-1}}.\] Tangent line method ( TL): It suffices to prove that $$\frac{x_i}{\sqrt{1 - x_i}} - \frac{\sqrt{x_i}}{\sqrt{n-1}} \ge \frac{1}{2(1 - 1/n)^{3/2}}(x_i - 1/n).$$(The desired result follows by summing cyclically.) (Note: RHS is the tangent line of LHS at $x_i = 1/n$.)