I want to show that $$\textbf{ln}(\sum_{i=1}^ne^{x_i})$$ for $x = (x_1, ..., x_n)^T \in \mathbb{R}^n$ is convex. I know that to do this, I can prove that its hessian is positive semidefinite.
I obtain that $$y^THy = \frac{1} {J}\sum_{i=1}^n y_i^2e^{x_i} - (\frac{1}{J}\sum_{i=1}^n y_ie^{x_i})^2,$$ where $J = \sum_{j=1}^n e^{x_j}$ and $y_i$ are the entries in the vector $y$ and $H$ is the hessian.
I'm trying to bound the quantity above using Cauchy Schwarz but I haven't been able to do this so far. Is it possible? Thanks for any help you can provide!
Note that $$J^2y^Thy=\sum_i{(e^{x_i/2})^2}\sum_i{(y_ie^{x_i/2})^2}-\left(\sum_i{y_ie^{x_i/2}e^{x_i/2}}\right)^2.$$