Suppose we have a row vector $$X = [x_1\; \cdots \; x_n],$$ where $x_i>0$ for all $i$. I have the function \begin{equation} \frac{d}{dk}\left(\frac{x_{i}^{k}}{\sum_{j}x_{j}^{k}}\right).\quad (1) \end{equation} I'm trying to prove that (1) is negative for all $x_i\neq \max\{x_j\}$. This seems somewhat intuitive to me. However, my attempts to `prove' it have not been successful.
The obvious approach is to use the quotient rule, i.e., \begin{align*} \frac{d}{dk}\left(\frac{x_{i}^{k}}{\sum_{j}x_{j}^{k}}\right)&=\frac{x_{i}^{k}}{\left(\sum_{j}x_{j}^{k}\right)^{2}}\sum_{j}\frac{dx_{j}^{k}}{dk}-\frac{dx_{i}^{k}}{dk}\frac{\sum_{j}x_{j}^{k}}{\left(\sum_{j}x_{j}^{k}\right)^{2}} , \end{align*} which could be carried forward as follows \begin{align*} \frac{d}{dk}\left(\frac{x_{i}^{k}}{\sum_{j}x_{j}^{k}}\right)&=\frac{x_{i}^{k}}{\left(\sum_{j}x_{j}^{k}\right)^{2}}\sum_{j}\frac{dx_{j}^{k}}{dk}-\frac{dx_{i}^{k}}{dk}\frac{\sum_{j}x_{j}^{k}}{\left(\sum_{j}x_{j}^{k}\right)^{2}} \\ & =\frac{x_{i}^{k}}{\left(\sum_{j}x_{j}^{k}\right)^{2}}\sum_{j}\frac{dx_{j}^{k}}{dk}-\frac{1}{\sum_{j}x_{j}^{k}}\frac{dx_{i}^{k}}{dk}\\ & =\frac{1}{\sum_{j}x_{j}^{k}}\left(\frac{x_{i}^{k}\sum_{j}\frac{dx_{j}^{k}}{dk}}{\sum_{j}x_{j}^{k}}-\frac{dx_{i}^{k}}{dk}\right)\\ & =\frac{1}{\sum_{j}x_{j}^{k}}\left(\frac{x_{i}^{k}\frac{d}{dk}\left(\sum_{j}x_{j}^{k}\right)}{\sum_{j}x_{j}^{k}}-\frac{dx_{i}^{k}}{dk}\right). \end{align*}
It's not clear to me whether this is the right track to be going down, or how the conclusion could be drawn from this. Can anyone help?
You want $R=x_i^k\big/\sum_j x_j^k$ to be decreasing in $k$. Look instead at $S=1/R$, which is given by $$S=\sum_j \left(\frac{x_j}{x_i}\right)^k.$$ This is manifestly a convex function of $k$, and if any of the ratios $x_j/x_i >1$, increasing for all $k$ sufficiently large.
Depending on whether $x_i$ is the largest, smallest, or neither, of the $x_j$, the graph of $k\mapsto S$ can look, qualitatively, like that of $\exp (-k)$, $\exp k$, or $\cosh k$.
So the answer depends in part on whether you suppose $k>0$ or not.