shobber wrote: Given that $x_{i}>0$, $i = 1, 2, \cdots, n$, $k \geq 1$. Show that: \[\sum_{i=1}^{n}\frac{1}{1+x_{i}}\cdot \sum_{i=1}^{n}x_{i}\leq \sum_{i=1}^{n}\frac{x_{i}^{k+1}}{1+x_{i}}\cdot \sum_{i=1}^{n}\frac{1}{x_{i}^{k}}\] $\sum_{i=1}^{n}\frac{x_{i}^{k+1}}{1+x_{i}}\cdot \sum_{i=1}^{n}\frac{1}{x_{i}^{k}}\geq \frac{1}{n}\sum_{i=1}^{n}x_{i}\cdot\ \sum_{i=1}^{n}\frac{x_{i}^{k}}{1+x_{i}}\cdot \sum_{i=1}^{n}\frac{1}{x_{i}^{k}}\geq \sum_{i=1}^{n}\frac{1}{1+x_{i}}\cdot \sum_{i=1}^{n}x_{i}$

Ummm...could you explain that a little more?

I don't quite follow either -- I assume that was Chebyshev applied twice, but the second inequality is going in the wrong direction? (Correct me if I'm wrong...) Anyways, here's my solution: Without loss of generality, $x_{1}\leq \cdots \leq x_{n}$. Then, $\sum_{i=1}^{n}\frac{x_{i}^{k+1}}{1+x_{i}}\sum_{i=1}^{n}\frac{1}{x_{i}^{k}}-\sum_{i=1}^{n}\frac{1}{1+x_{i}}\sum_{i=1}^{n}x_{i}$ $= \sum_{i>j}(x_{i}^{k+1}-x_{j}^{k+1}) \left( \frac{1}{(1+x_{i})x_{j}^{k}}-\frac{1}{(1+x_{j})x_{i}^{k}}\right) \geq 0$