This is from Boyd and Vanderberghe's Convex Optimization, exercise 3.17. I gave an abbreviated version in the question title: what I'd really like to know is how to prove that $\nabla^2f \preceq 0$ for the function $f:\mathbb{R}^n \to \mathbb{R}$ $$f(x) = \bigg(\sum_{i=1}^n x_i^p\bigg)^{1/p}$$ where $p<1$ and $p \neq 0$, so that I can prove that it's concave. I found that $$\frac{\partial f^2}{\partial x_1^2} = \frac{(1-p)}{p^2} \bigg(\sum_{i=1}^n x_i^p\bigg)^{(\frac{1}{p}-2)}$$
and that any second order derivative of $f$, and thus any element of $\nabla^2f$, is equal to this, since the expression treats all the $x_i$ identically.
However, this expression turns out to always be positive when $p<1$ and $p \neq 0$. So it proves the opposite of what the exercise asked me to prove! I feel like I must have missed a negative sign somewhere, but I can't find it for the life of me.