I'm trying to prove the convexity of a function from its Hessian matrix.
He = $\begin{bmatrix} \frac{1}{x_1} - \frac{1}{\sum_{i=1}^n x_i} & & \frac{-1}{\sum_{i=1}^n x_i} & \cdots & \frac{-1}{\sum_{i=1}^n x_i}\\ \frac{-1}{\sum_{i=1}^n x_i} & & \frac{1}{x_2} - \frac{1}{\sum_{i=1}^n x_i} & \cdots & \frac{-1}{\sum_{i=1}^n x_i}\\ \vdots & & \ddots & & \vdots \\ \frac{-1}{\sum_{i=1}^n x_i} & & \cdots & & \frac{1}{x_n} - \frac{1}{\sum_{i=1}^n x_i}\\ \end{bmatrix}$
My idea was to separate this into two matrices and see if we could get something out of that:
He = $\begin{bmatrix} \frac{1}{x_1} & 0 & 0 & \cdots & 0\\ 0 & \frac{1}{x_2} & 0 & \cdots & 0\\ \vdots & \ddots & \ddots & \ddots & 0\\ 0 & 0 & \cdots & 0 & \frac{1}{x_n}\\ \end{bmatrix}$ + $\begin{bmatrix} \frac{-1}{\sum_{i=1}^n x_i} & \frac{-1}{\sum_{i=1}^n x_i} & \cdots & \frac{-1}{\sum_{i=1}^n x_i}\\ \frac{-1}{\sum_{i=1}^n x_i} & \ddots & \ddots & \frac{-1}{\sum_{i=1}^n x_i}\\ \vdots & \ddots & \ddots & \vdots\\ \frac{-1}{\sum_{i=1}^n x_i} & \frac{-1}{\sum_{i=1}^n x_i} & \cdots & \frac{-1}{\sum_{i=1}^n x_i}\\ \end{bmatrix}$
Knowing that: $x \in \mathbb{R}_{++}^n$
we can see that the first matrix is positive definite and the second matrix consists of the same value in each position, and this value must be negative.
Also, since $\dfrac{1}{x_i} > \dfrac{1}{\sum_{i=1}^n x_i}$ for all $i$, the entries on the diagonal of the original Hessian must all be positive.
However, I don't know what else I can do with this.