Uniqueness of the solution of the functional equation $ \frac{\phi(x) + \phi(y)}{2} = \phi\left(\frac{2xy}{x+y}\right) $

Question

TL;DR. I'm trying to understand why the parameter $\beta$ in the Gibbs measure is the inverse of the temperature $\frac{1}{T}$ in thermal dynamical context.

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In the space of smooth bijections (diffeomorphisms) from $(0,\infty)$ to $(0,\infty)$, the function

$$ x \mapsto \frac{1}{x}$$

satisfies the functional equation

$$ \frac{\phi(x) + \phi(y)}{2} = \phi(\frac{2xy}{x+y}).$$

Indeed,

$$ \frac{\frac{1}{x} + \frac{1}{y}}{2} = \frac{x+y}{2xy}.$$

Question

Is this the only solution?

Attempts and motivation

I've used some techniques.. like investigating limits, finding special values, or differentiating $x\phi(x)$, etc. This question comes from statistical mechanics. It will help me understand, after accepting that the Gibbs measure

$$ \mu(s) \sim \exp(-\beta s) $$

is natural, why the parameter $\beta$ introduced from Lagrange multiplier method naturally corresponds to the inverse of the temperature $\frac{1}{T}$ in thermal dynamical context.

Answer 1

Hints for finding $\phi$: Differentiate w.r.t. $x$ to get $\frac 1 2 \phi'(x)=\phi '(\frac {2xy} {x+y}) \frac {2y^{2}} {(x+y)^{2}}$. Now put $x=1$ to get $ \phi '(\frac {2y} {1+y})$. Put $t=\frac {2y} {1+y}$ and you will get $\phi'(t)$ for every $t \in (0,1)$.

Treating $\phi'(1)$ as any given constant $c$, then you'll have

$$t^2 \phi'(t) = c.$$

By mean value theorem, $\phi(t) = c/t$ are the only solutions.