Is the geometric-to-arithmetic function convex or concave?

Question

Consider a vector $\mathbf{x} \in \mathbb{R}_{++}^N$. Also consider two functions, $g(\mathbf{x}): \mathbb{R}^N \rightarrow \mathbb{R}$, and $a(\mathbf{x}): \mathbb{R}^N \rightarrow \mathbb{R}$, representing the geometric and arithmetic means respectively.

The geometric mean function, $g(\mathbf{x})$, is concave. Furthermore, the arithmetic mean function, $a(\mathbf{x})$, is (I think), convex.

We also know that $0 \leq \frac{g(\mathbf{x})}{a(\mathbf{x})} \leq 1$ is always true.

My question is: Is the ratio $h(\mathbf{x}) = \frac{g(\mathbf{x})}{a(\mathbf{x})}$, convex or concave for this specific case?

I am currently trying to prove it by brute force on paper by deriving the Hessian matrix of $h(\mathbf{x})$, and then deriving its definiteness, but I was wondering if there might be an easier way.

Thanks.

Answer 1

It is neither convex nor concave, but it is quasiconcave. See my answer to your other question for more information, or go straight to Boyd & Vandenberghe.

The good news is that fixed lower bounds on the ratio can be represented in convex optimization problems; since, after all, $$g(x)/a(x) \geq \alpha \quad\Longleftrightarrow\quad g(x) \geq \alpha a(x).$$ (This depends upon your claim that $x\in\mathbb{R}^n_{++}$, so $a(x)>0$). You can even maximize the ratio by solving a sequence of convex optimization problems.

The arithmetic mean, incidentally, is both convex and concave; that is, it is affine (indeed, linear). So unlike the more general answer I gave on the other page, the constraint conversion works even if $\alpha<0$. Of course, if $\alpha<0$, it is trivially satisfied.

Answer 2

If $x=(x_1,\dots,x_n)\in\mathbb{R}^n$, $x_k>0$, then $$ \frac{g(x)}{f(x)}=n\frac{(x_1\dotsm x_n)^{1/n}}{x_1+\dots+x_n}. $$ Fix $x_2=\dots=x_n=1$. It is easy to see that the resulting function $$ n\frac{x_1^{1/n}}{x_1+n-1} $$ is neither concave nor convex, implying the same for $g(x)/a(x)$.

Graph of $h(x,y)$ for $0<x<1$ and $0<y<10$: