Given $m \in (0,1)$ and $n \in \mathbb{N}$, find a symmetric differentiable function $f : (0, \infty)^n \rightarrow \mathbb{R}$ that satisfies $$ \frac{\partial f(x)}{\partial x_i} = 0 \Longleftrightarrow \left(\sum_{i} x_i^m \right)^2 - m x_i^{m-1} \left( \sum_{j \neq i} x_j^m \right) = 0, $$ for every $i \in [n]$ and $x = (x_1, x_2, \ldots, x_n) \in (0,\infty)^n$.
I am not sure how to systematically approach this problem. Here are my current thoughts:
- The second term in the partial derivative, $-m x_i^{m-1} \left( \sum_{j \neq i} x_j^m \right)$, can be achieved using the following term in the function: $-\sum_{i < j} x_i^m x_j^m$.
- But, if we have a term $ \frac{\left(\sum_{i} x_i^m \right)^3}{3m}$ in the function, then we get $x_i^{m-1} \left(\sum_{i} x_i^m \right)^2$ and not $\left(\sum_{i} x_i^m \right)^2$ as required.
- The condition $\left(\sum_{i} x_i^m \right)^2 - m x_i^{m-1} \left( \sum_{j \neq i} x_j^m \right) = 0$ is equivalent to $ \left( \frac{m x_i^{m-1} \left( \sum_{j \neq i} x_j^m \right)}{\left(\sum_{i} x_i^m \right)^2} \right)^{\alpha} = 1,$ for any constant $\alpha \neq 0$. Other transformations can also be applied to get equivalent conditions, e.g., $ \psi\left( \frac{m x_i^{m-1} \left( \sum_{j \neq i} x_j^m \right)}{\left(\sum_{i} x_i^m \right)^2} \right) = \psi(1)$ for an injective function $\psi$.