[Largely corrected by following comments] Let $I$ be an open interval. Let $f : I \to \mathbb{R}$ be a strictly convex function in $C^2 (I)$. Let $$ d(x,y) := \sqrt{\frac{f(x)+f(y)}{2} - f\left(\frac{x+y}{2}\right)}, \ x, y \in I.$$ Then, I would like to know whether $d$ is a metric on $I$ or not.
The difficult part is the triangle inequality. In order to show the triangle inequality $d(x,z) \le d(x,y)+d(y,z)$, we let $F(y) := d(x,y)+d(y,z)-d(x,z)$, $y \in [x,z]$. (We assumed that $x < y < z$.) Then, $F(x) = F(z) = 0$. Therefore, I have tried to show that $F^{\prime\prime}(y) \le 0$, $y \in (x,z)$, by direct computations, but not yet succeeded.
Another approach I tried is to give an appropriate integral expression and use it. An example is as follows, but there are different expressions. $$ \frac{f(x)+f(y)}{2} - f\left(\frac{x+y}{2}\right) = \int_x^y \left| \int_{(x+y)/2}^{t} f^{\prime\prime}(u) du \right| dt, \ x < y. $$
An easy example is that if $f(x) = x^2, x \in \mathbb{R}$, then, $d(x,y) = |x-y|/2$, where $| \cdot |$ is the Euclid distance. In particular, I would like to know the case of $f(x) = x\log x, x > 0$. The background is Fuglede-Topsoe's paper for the Jensen-Shannon divergence.
I gave a thought to your question. I think it is not a distance. Possible counter example: Take $I=(0,\infty)$, $f(x)=x^3$ which is strictly convex on $I$. Then, $$ d(x,y)=\frac{\sqrt{3}}{2\sqrt{2}} \sqrt{x+y}\,|x-y|. $$ Now, $$ d(2,3)+d(3,5)-d(2,5)=\frac{\sqrt{3}}{2\sqrt{2}}\left(\sqrt{5}+4\sqrt{2}-3\sqrt{7}\right)<0 $$ which contradict the triangle inequality.