Interesting logarithmic inequality

Question

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.

Answer 1

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).

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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).

Answer 2

My attempt :

$\sqrt{ln (\frac{1+(\frac{b}{a}) ^2}{2\frac{b} {a} })}<\sqrt{ln (\frac{1+(\frac{c}{a}) ^2}{2\frac{c} {a} })}+ \sqrt{ln (\frac{1+(\frac{c}{b}) ^2}{2\frac{c} {b} })} $ Suppose $x=\frac{b}{a} $ and $y=\frac{c} {a} $

$\sqrt{ln (\frac{1+x^2}{2x})}<\sqrt{ln (\frac{1+y^2}{2y})}+\sqrt{ln (\frac{x^2+y^2}{2xy})}$

$\Leftrightarrow$$ln (\frac{1+x^2}{2x})<ln (\frac{1+y^2}{2y})+ln (\frac{x^2+y^2}{2xy})+2\sqrt{ln (\frac{1+y^2}{2y})×ln (\frac{x^2+y^2}{2xy})} $

$\Leftrightarrow$$ln (\frac{1+x^2}{2x})-ln (\frac{1+y^2}{2y})-ln (\frac{x^2+y^2}{2xy})-2\sqrt{ln (\frac{1+y^2}{2y})×ln (\frac{x^2+y^2}{2xy})}<0 $ $\Leftrightarrow$$ln (\frac{2y(1+x^2)}{2x(1+y^2})-ln (\frac{x^2+y^2}{2xy})-2\sqrt{ln (\frac{1+y^2}{2y})×ln (\frac{x^2+y^2}{2xy})}<0 $ $\Leftrightarrow$$ln (\frac{2y(1+x^2)(2xy) }{2x(1+y^2)(x^2+y^2)}-2\sqrt{ln (\frac{1+y^2}{2y})×ln (\frac{x^2+y^2}{2xy})}<0 $ $\Leftrightarrow$$ln (\frac{2y^2(1+x^2) }{(1+y^2)(x^2+y^2)}-2\sqrt{ln (\frac{1+y^2}{2y})×ln (\frac{x^2+y^2}{2xy})}<0 $ $\Leftrightarrow$$ln (\frac{(1+x^2)2y^2 }{(1+y^2)(x^2+y^2)}-2\sqrt{ln (\frac{1+y^2}{2y})×ln (\frac{x^2+y^2}{2xy})}<0 $

So that is always correct if $ ln(\frac{(1+x^2)2y^2 }{(1+y^2)(x^2+y^2)}$<0 $\Rightarrow$$\frac{(1+x^2)2y^2 }{(1+y^2)(x^2+y^2)}<1$

$\Rightarrow$$y^2x^2+y^2-x^2-y^4<0$$\Rightarrow$$ (x^2-y^2)(y^2-1)<0$$\Rightarrow$$ x<y<1$ or $x>y>1$

that is correct if

$b<c<1$ or $b>c>1 $

So I am showed this inequality for $b<c<1 $ and $b>c>1 $