I found a question, but was unable to solve it. The question is:
Let k be a positive integer, and let $x_1, x_2, ..., x_n$ be positive real numbers. Prove that $$\left(\sum_{i=1}^n \frac{1}{1+x_i}\right)\left(\sum_{i=1}^n x_i\right)\le\left(\sum_{i=1}^n \frac{x_i^{k+1}}{1+x_i}\right)\left(\sum_{i=1}^n \frac{1}{x_i^k}\right) .$$ I tried if I could somehow use rearrangement inequality but due to the powers, I was unable to apply it. Please suggest some method to solve this.
On
By Muirhead we obtain: $$\sum \frac{x_i^{k+1}}{1+x_i}\sum\frac{1}{x_i^k}-\sum\frac{1}{1+x_i}\sum x_i=$$ $$=\frac{1}{4}\sum_{i\neq j}\left(\frac{x_i^{k+1}}{(1+x_i)x_j^k}+\frac{x_j^{k+1}}{(1+x_j)x_i^k}-\frac{x_i}{1+x_j}-\frac{x_j}{1+x_i}\right)=$$ $$=\frac{1}{4}\sum_{i\neq j}\frac{x_i^{2k+1}(1+x_j)+x_j^{2k+1}(1+x_i)-x_i^kx_j^k(x_i+x_j+x_i^2+x_j^2)}{(1+x_i)(1+x_j)x_i^kx_j^k}=$$ $$=\frac{1}{4}\sum_{i\neq j}\tfrac{\left(x_i^{2k+1}+x_j^{2k+1}-x_i^{k+1}x_j^k-x_i^kx_j^{k+1}\right)+\left(x_i^{2k+1}x_j+x_j^{2k+1}x_i-x_i^{k+2}x_j^k-x_i^kx_j^{k+2}\right)}{(1+x_i)(1+x_j)x_i^kx_j^k}\geq0.$$
First, without loss of generality, let us renumber the reals in ascending order, i.e. $x_0 \le x_1 \le \cdots \le x_n$.
Then, write the question as $$ \sum_{i,j=0}^n \frac{x_j}{1+x_i}\le \sum_{i,j=0}^n \frac{x_j}{1+x_i} \Big[\frac{x_i}{x_j}\Big]^{k+1} $$ Let us call $w_{i,j} = \Big[\frac{x_i}{x_j}\Big]^{k+1}$. The diagonal terms equal, so this can be written $$ \sum_{i>j} (\frac{x_j}{1+x_i} + \frac{x_i}{1+x_j})\le \sum_{i>j} (w_{i,j} \frac{x_j}{1+x_i} + \frac{1}{w_{i,j}}\frac{x_i}{1+x_j}) $$ or $$ \sum_{i>j} \frac{ x_j (1+x_j) + x_i (1+x_i)}{(1+x_i)(1+x_j)} \le \sum_{i>j} \frac{w_{i,j} x_j (1+x_j) + \frac{1}{w_{i,j}} x_i (1+x_i)}{(1+x_i)(1+x_j)} $$ We can now inspect this for each term in the sum. If we can show the inequality for each term we are done. This reads: $$x_j (1+x_j) + x_i (1+x_i) \le w_{i,j} x_j (1+x_j) + \frac{1}{w_{i,j}} x_i (1+x_i)$$ which again is true if the two inequalities hold: $$x_j + x_i \le w_{i,j} x_j + \frac{1}{w_{i,j}} x_i \\ x_j^2 + x_i^2 \le w_{i,j} x_j^2 + \frac{1}{w_{i,j}} x_i^2 $$ Now these inequalities are of the form $a + b \le a w + b \frac1w $. When do they hold? Some simple equivalences: let $c = b/a \ge 1$ (see below that indeed $c \ge 1$), then $$ a + b \le a w + b \frac1w \\ \leftrightarrow c + 1\le w + c \frac1w\\ \leftrightarrow 0 \le w^2 - (c + 1)w + c\\ \leftrightarrow 0 \le (w- \frac12 (c + 1))^2 - \frac14 (c + 1)^2 + c\\ \leftrightarrow \frac14 (c - 1)^2 \le (w- \frac12 (c + 1))^2 \\ \leftrightarrow \frac12 (c - 1) + \frac12 (c + 1) \le w \\ \leftrightarrow c \le w \\ $$ Hence we need, reinstating the variables: $$x_j + x_i \le w_{i,j} x_j + \frac{1}{w_{i,j}} x_i \\ \leftrightarrow c = \frac{x_i}{x_j} \le w_{i,j} = \Big[\frac{x_i}{x_j}\Big]^{k+1} $$ which is true for all $k \ge 0$, since $c\ge 1 $ or $x_i \ge x_j$, as we consider the sum $\sum_{i>j} $ and the reals are numbered in ascending order.
Likewise, the second inequality is: $$x_j^2 + x_i^2 \le w_{i,j} x_j^2 + \frac{1}{w_{i,j}} x_i^2 \\ \leftrightarrow \frac{x_i^2}{x_j^2} \le w_{i,j} = \Big[\frac{x_i}{x_j}\Big]^{k+1} $$ which is true for all $k \ge 1$. Hence $k \ge 1$ is the stricter condition, which is given since the $k$ are positive integers. $ \qquad \Box$