Zhaobin,To prove this problem ,I only need to prove: ${\sum_{i=1}^n \frac{1}{\sqrt{n}+\sqrt{x_{i}}} \leq \frac{n^2}{n+1}\frac{1}{\sum_{i=1}^n \sqrt{x_i}}}$ when $x_{1}+x_{2}+....+x_{n}=1$ (1) (because $(\sum_{i=1}^n \frac{1}{\sqrt{1+x_{i}}} \leq \sqrt{n+1}(\sum_{i=1}^n \frac{1}{\sqrt{n}+\sqrt{x_{i}}})$) with (1),Now I have a solution but it's too long,So i'm confused to post it.I promise I'll post the solution to (1) if everyone can't find the solution to it

if $\sum tan^2 a_i=1$ then maybe we can show that $\sum sin a_i\le \frac{n}{\sqrt{n+1}}$

let $1 + x_i = y_i \Rightarrow x_i = y_i - 1,{\rm{ }}y_i > 1(i = 1,2, \cdots ,n),\sum\limits_{i = 1}^n {y_i } = n + 1$,the inequality is equal to:\[ \left( {\sum\limits_{i = 1}^n {\sqrt{y_i - 1} } } \right)\sum\limits_{i = 1}^n {\frac{1}{{\sqrt{y_i } }}} \le \frac{{n^2 }}{{\sqrt{n + 1} }}{\rm{ }} ................. (2) \] By Cauchy inequality \[ \begin{array}{l} \frac{1}{{\sqrt n }} \cdot \sqrt{y_1 - 1} + \sqrt{y_2 - 1} \cdot \frac{1}{{\sqrt n }} + \sqrt{y_3 - 1} \cdot \frac{1}{{\sqrt n }} + \cdots + \sqrt{y_n - 1} \cdot \frac{1}{{\sqrt n }} \\ \le \sqrt{[\frac{1}{n} + y_2 + y_3 + \cdots + y_n - (n - 1)](y_1 - 1 + \frac{{n - 1}}{n})} \\ = \sqrt{(\frac{1}{n} + n + 1 - y_1 - n + 1)(y_1 - \frac{1}{n})} = \sqrt{ - y_1 ^2 + \frac{{2(n + 1)}}{n}y_1 - \frac{{2n + 1}}{{n^2 }}} \\ \Rightarrow \frac{{\sum\limits_{i = 1}^n {\sqrt{y_i - 1} } }}{{\sqrt{y_1 } }} \le \sqrt n \sqrt{ - y_1 + \frac{2}{n}(n + 1) - \frac{{2n + 1}}{{n^2 }} \cdot \frac{1}{{y_1 }}} \\ \end{array} \] similarly \[ \begin{array}{l} \frac{{\sum\limits_{i = 1}^n {\sqrt{y_i - 1} } }}{{\sqrt{y_2 } }} \le \sqrt n \sqrt{ - y_2 + \frac{2}{n}(n + 1) - \frac{{2n + 1}}{{n^2 }} \cdot \frac{1}{{y_2 }}} \\ \cdots \\ \frac{{\sum\limits_{i = 1}^n {\sqrt{y_i - 1} } }}{{\sqrt{y_n } }} \le \sqrt n \sqrt{ - y_n + \frac{2}{n}(n + 1) - \frac{{2n + 1}}{{n^2 }} \cdot \frac{1}{{y_n }}} \\ \end{array} \] so that\[ \[ \begin{array}{l} {\rm{ }}\left( {\sum\limits_{i = 1}^n {\sqrt{y_i - 1} } } \right)\sum\limits_{i = 1}^n {\frac{1}{{\sqrt{y_i } }}} \le \sqrt n \sum\limits_{i = 1}^n {\sqrt{ - y_i + \frac{2}{n}(n + 1) - \frac{{2n + 1}}{{n^2 }} \cdot \frac{1}{{1 + y_i }}} } \\ \overbrace \le ^{{\rm{Cauchy}}}\sqrt n \cdot \sqrt n \sqrt{\sum\limits_{i = 1}^n {[ - y_i + \frac{2}{n}(n + 1) - \frac{{2n + 1}}{{n^2 }} \cdot \frac{1}{{1 + y_i }}} } ] \\ = n\sqrt{ - (n + 1) + 2(n + 1) - \frac{{2n + 1}}{{n^2 }}\sum\limits_{i = 1}^n {\frac{1}{{1 + y_i }}} } \\ = n\sqrt{n + 1 - \frac{{(2n + 1)}}{{n^2 }}\sum\limits_{i = 1}^n {\frac{1}{{y_i }}} } \\ \end{array} \] By Cauchy inequality \[ \sum\limits_{i = 1}^n {\frac{1}{{y_i }}} \ge \frac{{n^2 }}{{\sum\limits_{i = 1}^n {y_i } }} = \frac{{n^2 }}{{n + 1}} \] therefore \[ \left( {\sum\limits_{i = 1}^n {\sqrt{y_i - 1} } } \right)\sum\limits_{i = 1}^n {\frac{1}{{\sqrt{y_i } }}} \le n\sqrt{n + 1 - \frac{{(2n + 1)}}{{n^2 }}\sum\limits_{i = 1}^n {\frac{1}{{y_i }}} } \le n\sqrt{n + 1 - \frac{{2n + 1}}{{n^2 }} \cdot \frac{{n^2 }}{{n + 1}}} = \frac{{n^2 }}{{\sqrt{n + 1} }} \]

General Let \[ x_i > 0(i = 1,2, \cdots ,n),2 \le n \in N,\sum\limits_{i = 1}^n {x_i } = 1,2 \le m \in N,0 < \lambda \le \frac{n}{{n - 1}} \] that\[ (\sum\limits_{i = 1}^n {\sqrt[m]{{x_i }}} )\sum\limits_{i = 1}^n {\frac{1}{{\sqrt[m]{{1 + \lambda x_i }}}}} \le \frac{{n^2 }}{{\sqrt[m]{{n + \lambda }}}} ..........(3) \] with (1),Now I have a solution but it's too long。

thank you to you all. and thanks to scpajmb's nice solution. to nhat:can you post solution for (1),because it is also a nice inequality. to Albanian Eagle:this may be the offical solution:but I think it is not only $\sum sin a_i$

I think we can use AM-GM for the original inequality. It suffices to prove $n \sum \sqrt{x_i}+ \sqrt{n+1} \sum \frac 1{\sqrt{1+x_i}} \leq 2n\sqrt n$ and then apply AM-GM. This inequality follows from the concavity of the function $n \sqrt x+\sqrt{n+1} \frac 1{\sqrt{1+x}}$ on $(0;1)$

@scpajmb: I think, your solution isn't true, cause x is not x+1. felix

scpajmb wrote: so that\[ \[ \begin{array}{l} {\rm{ }}\left( {\sum\limits_{i = 1}^n {\sqrt{y_i - 1} } } \right)\sum\limits_{i = 1}^n {\frac{1}{{\sqrt{y_i } }}} \le \sqrt n \sum\limits_{i = 1}^n {\sqrt{ - y_i + \frac{2}{n}(n + 1) - \frac{{2n + 1}}{{n^2 }} \cdot \frac{1}{{1 + y_i }}} } \\ \overbrace \le ^{{\rm{Cauchy}}}\sqrt n \cdot \sqrt n \sqrt{\sum\limits_{i = 1}^n {[ - y_i + \frac{2}{n}(n + 1) - \frac{{2n + 1}}{{n^2 }} \cdot \frac{1}{{1 + y_i }}} } ] \\ = n\sqrt{ - (n + 1) + 2(n + 1) - \frac{{2n + 1}}{{n^2 }}\sum\limits_{i = 1}^n {\frac{1}{{1 + y_i }}} } \\ = n\sqrt{n + 1 - \frac{{(2n + 1)}}{{n^2 }}\sum\limits_{i = 1}^n {\frac{1}{{y_i }}} } \\ \end{array} \] in this part, all the $\frac{1}{1+y_i}$ should be $\frac{1}{y_i}$ then the others is no problem.

zhaobin wrote: $x_{1}, x_{2}, \cdots, x_{n}$ are positive numbers such that $\sum_{i=1}^{n}x_{i}= 1$. Prove that \[\left( \sum_{i=1}^{n}\sqrt{x_{i}}\right) \left( \sum_{i=1}^{n}\frac{1}{\sqrt{1+x_{i}}}\right) \leq \frac{n^{2}}{\sqrt{n+1}}\] I think it's very easy by using Cauchy-Schwartz Inequality : First, we have if $x_i \in [0,1]$ then : \[\sum_{i=1}^n \dfrac{1}{\sqrt{1+x_i}} \leq \dfrac{n}{\sqrt{1+\sqrt[n]{x_1x_2...x_n}}}\] So we only prove that: \[\dfrac{\left(\sqrt{x_1}+...+\sqrt{x_n}\right)^2}{1+\sqrt[n]{x_1x_2...x_n}}\leq \dfrac{n^2}{n+1}\] \[\Longleftrightarrow \[\dfrac{\left(\sqrt{x_1}+...+\sqrt{x_n}\right)^2}{x_1+x_2+...+x_n+\sqrt[n]{x_1x_2...x_n}}\leq \dfrac{n^2}{n+1}\] Assume that $x_1x_2...x_n=1$ we need to prove: \[\dfrac{\left(\sqrt{x_1}+...+\sqrt{x_n}\right)^2}{x_1+x_2+...+x_n+1}\leq \dfrac{n^2}{n+1}\] By Cauchy-Schwartz Inequality, we have: \[\dfrac{\left(\sqrt{x_1}+...+\sqrt{x_n}\right)^2}{x_1+x_2+...+x_n+1}\leq \sum_{i=1}^n \dfrac{x_i}{x_i+\dfrac{1}{n}}\] So we need to prove: \[ \sum_{i=1}^n \dfrac{nx_i}{nx_i+1}\leq \dfrac{n^2}{n+1}\] \[\Longleftrightarrow \sum_{i=1}^n \dfrac {1}{nx_i+1}\geq \dfrac{n}{n+1}\] Let $a_i=\dfrac{1}{x_i}$ we have $a_1a_2...a_n=1$ and we need to prove: \[\sum_{i=1}^n\dfrac{a_i}{a_i+n}\geq \dfrac{n}{n+1}\] Using Cauchy-Schwartz inequality, we have: \[\sum_{i=1}^n\dfrac{a_i}{a_i+n}\geq \dfrac{\left(a_1+...+a_n\right)^2}{a_1^2+...+a_n^2+n^2}\] We prove that : \[ \dfrac{\left(a_1+...+a_n\right)^2}{a_1^2+...+a_n^2+n^2} \geq \dfrac{n}{n+1}\] \[\Longleftrightarrow \left(a_1+...+a_n\right)^2 +2n \sum_{i<j}a_ia_j \geq n^3\] It is true because: \[ \left\{ \begin{array}{ll} \left(a_1+...+a_n\right)^2\geq n^2\\ \sum_{i<j}a_ia_j \geq \binom{n}{2}=\dfrac{n(n-1)}{2} \end{array} \right.\] \[Q.E.D\]

ndk09 wrote: zhaobin wrote: $x_{1}, x_{2}, \cdots, x_{n}$ are positive numbers such that $\sum_{i=1}^{n}x_{i}= 1$. Prove that \[\left( \sum_{i=1}^{n}\sqrt{x_{i}}\right) \left( \sum_{i=1}^{n}\frac{1}{\sqrt{1+x_{i}}}\right) \leq \frac{n^{2}}{\sqrt{n+1}}\] I think it's very easy by using Cauchy-Schwartz Inequality : First, we have if $x_i \in [0,1]$ then : \[\sum_{i=1}^n \dfrac{1}{\sqrt{1+x_i}} \leq \dfrac{n}{\sqrt{1+\sqrt[n]{x_1x_2...x_n}}}\] So we only prove that: ... \[ \left\{ \begin{array}{ll} \left(a_1+...+a_n\right)^2\geq n^2\\ \sum_{i<j}a_ia_j \geq \binom{n}{2}=\dfrac{n(n-1)}{2} \end{array} \right.\] \[Q.E.D\] Nice proof, it is quite similar to mine. Thanks; ndk. By the way, I have posted my proof here.

Here is a very nice solution: By the Cauchy-Schwarz Inequality, we have \begin{align*} \sqrt{x_1} + \sqrt{x_2} + \cdots + \sqrt{x_n} &\le \sqrt{\bigg((1 + x_1) + (1 + x_2) + \cdots + (1 + x_n)\bigg)\left(\frac{x_1}{1 + x_1} + \frac{x_2}{1 + x_2} + \cdots + \frac{x_n}{1 + x_n}\right)} \\ &= \sqrt{(n + 1)\left(\frac{(1 + x_1) - 1}{1 + x_1} + \frac{(1 + x_2) - 1}{1 + x_2} + \cdots + \frac{(1 + x_n) - 1}{1 + x_n}\right)} \\ &= \sqrt{(n + 1)\left(n - \frac{1}{1 + x_1} - \frac{1}{1 + x_2} + \cdots - \frac{1}{1 + x_n}\right)}. \end{align*} Also by the Cauchy-Schwarz Inequality, we have \begin{align*} \frac{1}{\sqrt{1 + x_1}} + \frac{1}{\sqrt{1 + x_2}} + \cdots + \frac{1}{\sqrt{1 + x_n}} &\le \sqrt{\left(1 + 1 + \cdots 1\right)\left(\frac{1}{1 + x_2} + \frac{1}{1 + x_2} + \cdots + \frac{1}{1 + x_n}\right)} \\ &= \sqrt{n\left(\frac{1}{1 + x_2} + \frac{1}{1 + x_2} + \cdots + \frac{1}{1 + x_n}\right)}. \end{align*} Therefore, if we set $t = \sum\limits_{i = 1}^{n} \frac{1}{1 + x_i}$, it suffices to prove that $\sqrt{(n + 1)(n - t)} \cdot \sqrt{nt} \le \frac{n^2}{\sqrt{n + 1}}.$ Upon squaring and rearranging, this is just \[(n - t)t \le \frac{n^3}{\left(n + 1\right)^2} \iff \left(t - \frac{n^2}{n + 1}\right)^2 + \frac{n(n - 1)}{n + 1}\left(t - \frac{n^2}{n + 1}\right) \ge 0,\] where we have completed the square above. Hence, it suffices to prove that $t - \frac{n^2}{n + 1} \ge 0$, but this follows immediately from the convexity of the continuous and differentiable function $f : (0, 1) \to \mathbb{R}^+$ defined by $f(t) = \frac{1}{1 + t}.$ We easily compute $f''(t) = \frac{2}{\left(1 + t\right)^3} \ge 0,$ so $f$ is convex. Therefore, by Jensen's Inequality, we have \[\frac{1}{1 + x_1} + \frac{1}{1 + x_2} + \cdots + \frac{1}{1 + x_n} \ge n \cdot \frac{1}{1 + \frac{x_1 + x_2 + \cdots + x_n}{n}} = \frac{n^2}{n + 1},\] as desired. $\square$

Jia-Jia-Jia-Jia wrote:

There is something wrong in your using of Chebyshev. Maybe Chebyshev Inequality can't be used in this problem.