<s>[INST] Interesting logarithmic inequality 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.
 [/INST] 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 $
</s>