mr.danh wrote: Let $ x_1, x_2, ... , x_n$ be positive real numbers such that $ x_1x_2 ... x_n = 1$. Prove that $ \sum_{i = 1}^n \frac {1}{n - 1 + x_i}\le 1$ uhm,here is also the ineq in Romania TST 1998,and AMGM works here

this is also from Romania 1999 I think: with contradition method,assume that: $ \sum_{i = 1}^n\frac 1{n - 1 + x_i} > 1$ hence: \begin{eqnarray*}\frac 1{n - 1 + x_1} & > & 1 - \left(\sum_{i = 2}^n\frac 1{n - 1 + x_i}\right) \\ & = & \sum_{i = 2}^n\left( \frac 1{n - 1} - \frac 1{n - 1 + x_2}\right) \\ & = & \sum_{i = 2}^n\frac {x_i}{(n - 1)(n - 1 + x_i)} \\ & \geq & (n - 1)\sqrt [n - 1]{\frac {\prod_{i = 2}^n x_i}{(n - 1)^{n - 1}\prod_{i = 2}^n (n - 1 + x_i)}} \\ & = & \sqrt [n - 1]{\frac {\prod_{i = 2}^n x_i}{\prod_{i = 2}^n (n - 1 + x_i)}}\end{eqnarray*} (the inequality part is deduced by AM-GM) hence: $ \frac 1{n - 1 + x_1} > \sqrt [n - 1]{\frac {\prod_{i = 2}^n x_i}{\prod_{i = 2}^n (n - 1 + x_i)}}$ now write similar inequalities for other $ x_i$s and multiple them,we get that: $ x_1x_2\ldots x_n < 1$ which is a contradiction.

mr.danh wrote: Let $ x_1, x_2, ... , x_n$ be positive real numbers such that $ x_1x_2 ... x_n = 1$. Prove that $ \sum_{i = 1}^n \frac {1}{n - 1 + x_i}\le 1$ And the Cauchy Schwarz helps here, too. The inequality equalivents with \[ \sum {\frac{a}{{n - 1 + a}}} \ge 1 \] By the Cauchy Schwarz we have \[ \sum {\frac{a}{{n - 1 + a}}} \ge \frac{{\left( {\sum a } \right)^2 }}{{n\left( {n - 1} \right) + \sum a }} \] And $ \left( {\sum a } \right)^2 \ge n\left( {n - 1} \right) + \sum a \Leftrightarrow \left( {\sum a - n} \right)\left( {\sum a + n - 1} \right)$, it's done, .

Honey_S wrote: By the Cauchy Schwarz we have \[ \sum {\frac {a}{{n - 1 + a}}} \ge \frac {{\left( {\sum a } \right)^2 }}{{n\left( {n - 1} \right) + \sum a }} \] I don't think so,by cauchy schwarz we get that: $ \sum\frac a{n-1+a}\geq\frac{\left(\sum a\right)^2}{\sum a^2+(n-1)\sum a}$ which doesn't help us at all...

Honey_8 your solution is not right，because you forget their products is 1。

Can we use Jensen's inequality for this? Because the function is concave, so $\sum_{i = 1}^{n}\frac{1}{n-1+x_{i}} \le n $ $\times$ $\frac{1}{n-1+\frac{x_1+x_2+\dots+x_n}{n}}$ $\le$ $\frac{n}{n-1+1} = 1$ I'm not sure if I'm right though, please correct me if I'm wrong, thanks!

jack2627 wrote: Can we use Jensen's inequality for this? Because the function is concave, so $\sum_{i = 1}^{n}\frac{1}{n-1+x_{i}} \le n $ $\times$ $\frac{1}{n-1+\frac{x_1+x_2+\dots+x_n}{n}}$ $\le$ $\frac{n}{n-1+1} = 1$ I'm not sure if I'm right though, please correct me if I'm wrong, thanks! It's wrong because $f(x)=\frac{1}{n-1+x}$ is a convex function.

See also here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=238609

Singapore Mathematical Olympiad 2004 , Special Round: Let $ x_1, x_2,\ldots , x_n\in(0,1]$ and $n\ge 1$. Prove that \[\sum_{i = 1}^n \frac { x_i}{1+(n - 1 ) x_i}\le 1.\]

Sqing's problem can be solved by BaBak Ghalebi's approach.