Is there an analogue of convexity for geometric mean?

Question

Jensen's inequality implies that $$f(x/2 + y/2) \leq f(x)/2 + f(y)/2$$ for convex $f$, and the inequality reverses if $f$ is concave. Is there a class of functions such that similar inequalities hold for geometric means, that is, $$f(\sqrt{xy}) \leq \sqrt{f(x)f(y)}$$ or $$f(\sqrt{xy}) \geq \sqrt{f(x)f(y)}?$$

Answer 1

Generally speaking, $f$ is convex iff $f(tx+(1-t)y)\leq tf(x)+(1-t)f(y)$ for all $0\leq t\leq 1$. An analogous condition for the geometric mean might be $f(x^ty^{1-t})\leq f(x)^tf(y)^{1-t}$ for $0\leq t\leq 1$. In this case, we clearly have $$f(\sqrt{xy}) = f(x^{1/2}y^{1/2})\leq f(x)^{1/2}f(y)^{1/2} = \sqrt{f(x)f(y)}$$ The opposite condition, corresponding to concavity in the geometric mean, would be $f(x^ty^{1-t})\geq f(x)^tf(y)^{1-t}$ for $0\leq t\leq 1$.

Some examples: Let's call our condition "exponential convexity," for want of a better term. An example of a function that is exponentially convex and exponentially concave (and therefore has zero exponential convexity) would be $x^p$ for $p\in \mathbb{R}$. We note that $$f(x^ty^{1-t}) = (x^t)^p(y^{1-t})^p = (x^p)^t(y^p)^{1-t} = f(x)^tf(y)^{1-t}$$ An example of a function that is strictly exponentially convex would be $e^x$ restricted to $x > 0$. We note that $$f(x^ty^{1-t}) = e^{x^ty^{1-t}} < e^{tx+(1-t)y} = f(x)^tf(y)^{1-t}$$ for $x\neq y$ and $0 < t < 1$ by the weighted AM-GM inequality. An example of a function that is strictly exponentially concave would therefore be $\log(x)$ restricted to $x > 1$. We note that $$f(x^ty^{1-t}) = \log(x^ty^{1-t}) = t\log(x)+(1-t)\log(y) > \log(x)^t\log(y)^{1-t} = f(x)^tf(y)^{1-t}$$ for $x\neq y$ and $0 < t < 1$, also by the weighted AM-GM inequality.