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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.

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    $\begingroup$ Erm... what kind of aggressive comment is that, @NoNames? You manage to implicitly cast doubt on the skills of the OP and call them lazy, this sounds like an amazing strategy to build a welcoming, inclusive website. $\endgroup$
    – Clement C.
    Mar 1 at 19:56
  • $\begingroup$ @Clement C. Obviously, you are new, here. It is expected to show own effort to solve, or at least research, the questions you ask, here. "Welcoming" is optional, and if somebody with a "reputation" of 18.5k asks, I doubt they need much welcome, they know the ropes. $\endgroup$
    – NoNames
    Mar 1 at 20:02
  • $\begingroup$ Hahaha. Yes, @NoNames, obviously I am new here. As my profile clearly indicates. $\endgroup$
    – Clement C.
    Mar 1 at 20:02
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    $\begingroup$ @NN2 You are mistaken. The equality holds if $c=a$ or $c=b$. $\endgroup$
    – user
    Mar 1 at 20:25
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    $\begingroup$ As a side comment: 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). $\endgroup$
    – Clement C.
    Mar 1 at 23:22
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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).

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    $\begingroup$ If the above is correct, the same proof generalizes to any power $\alpha\in[0,1/2]$ instead of $\sqrt{\cdot}$, and explains why this fails for $\alpha > 1/2$ (the resulting function $h$ will not be concave). $\endgroup$
    – Clement C.
    Mar 1 at 23:54
  • $\begingroup$ Excellent! It appears to be even nicer and deeper than I expected. $\endgroup$
    – user
    Mar 2 at 6:24
  • $\begingroup$ @user Glad to hear! Make sure to double-check what I wrote, one is never too careful :) $\endgroup$
    – Clement C.
    Mar 2 at 6:31
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    $\begingroup$ Of course I have checked before writing my comment. :) $\endgroup$
    – user
    Mar 2 at 6:33
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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 $

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