TITLE: Interesting logarithmic inequality
QUESTION [5 upvotes]: Recently the following question was published on the site and soon after this deleted:

Prove the inequality
$$
\left(\log\frac{a^2+b^2}{2ab}\right)^n\le \left(\log\frac{a^2+c^2}{2ac}\right)^n+\left(\log\frac{b^2+c^2}{2bc}\right)^n\tag1
$$
for all positive $a,b,c$ and $n=\frac12$.

The inequality has many nice features. Particularly $\frac12$ seems to be the largest possible value of $n$ such that (1) generally holds.
I have succeeded to prove the inequality in a rather awkward way but I believe there should be a simple and nice way to prove it. Possibly it is just a particular case of a more general inequality.
Any hint is appreciated.

REPLY [3 votes]: Letting $f\colon (0,\infty)\times(0,\infty)$ be defined as
$$
f(x,y) = \sqrt{\log \frac{x^2+y^2}{2xy}}
$$
you question is equivalent to showing that $f$ defines a metric. (This is because we do have, for all $x,y>0$, that $f(x,y)\geq 0$, $f(x,y)=f(y,x)$, and $f(x,y)=0$ iff $x=y$, already).

It remains to prove the triangle inequality, which is the focus of this question. To do so, let us rewrite, for $x,y>0$.
$$
f(x,y) = \sqrt{\log\left(\frac{\frac{x}{y} + \frac{y}{x}}{2}\right)} \tag{1}
$$
and, going further, let us define $h\colon\mathbb{R}\to\mathbb{R}$ by
$$
h(t) = \sqrt{\log\left(\frac{e^t + e^{-t}}{2}\right)} = \sqrt{\log\cosh t}, \qquad t\in\mathbb{R} \tag{2}
$$
so that
$
f(x,y) = h\!\left(\log\frac{x}{y}\right)
$ for all $x,y>0$. Now, the inequality we want to show is
$$
f(x,y) \leq f(x,z) + f(y,z), \qquad x,y,z>0\tag{3}
$$
for which it is sufficient to prove, if I am not mistaken, that
$$
h(s+t) \leq h(s)+h(t), \qquad s,t\in\mathbb{R}\tag{4}
$$
by setting $s=\log\frac{x}{z}$, $t=\log\frac{z}{y}$, so that $s+t=\log\frac{x}{y}$ (which was the whole point of introducing $h$: making things nice and additive).
This last inequality follows from the fact that $h$ is subadditive, because concave (one can differentiate twice, I assume, to see it, and deal with $0$ separately).