Inverse of $\sum^n_{i=1} 1/x_i$ convex?

Question

Is this function convex? $$f(x)=\left(\sum^n_{i=1} 1/x_i \right)^{-1}$$ where $\operatorname{dom} f =\mathbb{R}^n_{++}$.

I having a hard time coming up with a clean Hessian from the partial derivatives.

Answer 1

If I've not made any mistakes in my calculations: if $n \geq 2$, then

$$ \partial_1 f(x) = -\frac{1}{\left( \sum_{i=1}^n x_i^{-1} \right)^2} \cdot \left(- \frac{1}{x_1^2} \right) = \frac{1}{ \left(1 + \sum_{i \neq 1} \tfrac{x_1}{x_i} \right)^2 }, \\ \partial_1^2 f(x) = - \frac{2}{\left(1 + \sum_{i \neq 1} \tfrac{x_1}{x_i} \right)^3} \cdot \left( \sum_{i \neq 1} x_i^{-1} \right) < 0 \text{ on } \mathbb{R}_{++}^n $$ so all Hessians will have a negative entry in the upper left corner, and hence they can't be positive (semi-)definite. Hence $f$ is not convex for $n \geq 2$.

EDIT: It becomes even less mystical when considering restrictions of the function at specific values. Let $n = 2$ and $y = 1$, then $f(x,1) = \left(\tfrac{1}{x} +1 \right)^{-1}$ is easily plotted to not be convex.