How to prove that the functional equation $f(x)+f(y)=f(\frac{xf'(x)+yf'(y)}{f'(x)+f'(y)})+f(\frac{x+y}2)$ is verified only by some basic functions? It's an question about functional analysis :

Let $x,y>0$ and $f$ a continuous and convex/concave function on $(0,\infty)$ with $f'(x)\neq 0$ on the previous interval  and:
$$f(x)+f(y)=f\left(\frac{xf'(x)+yf'(y)}{f'(x)+f'(y)}\right)+f\left(\frac{x+y}{2}\right)$$

My claim is :

The only function verifying this equation are
$$\pm (cx+b)$$
$$\pm (\frac{c}{x}+b)$$
$$\pm \ln(x)$$

It's not hard to check that the functions above verify the functional equation .
Edit :
It seems that the only function wich verify the functional above are reciprocally convex/concave .
My question :
How to prove it ? Is the sentence of my edit true ?
Thanks a lot for your time .
 A: Answer:
Besides the families
$f(x)=ax+b$, $f(x)=\frac a{x+b}+c$ and $f(x)=a\ln(bx+c)$ of solutions of the
functional equation
$$f(x)+f(y)=f\left(\frac{xf'(x)+yf'(y)}{f'(x)+f'(y)}\right)+f(\frac{x+y}2)$$
mentioned in the question, there
also are the families
$$f(x)=a\ln\left(\frac{x+b}{x+c}\right)+d\mbox{ with positive }b,c$$
and
$$f(x)=a\arctan(bx+c)+d\mbox{ if }b\neq0.$$
This completes the list of all its solutions.
It turns out, that members of the first new family are also reciprocally convex/concave,
membres of the second one only if $bc\geq0$.
If $bc<0$, then the members of the second new family are not even convex/concave.
Observe that the only solutions of the functional equation that can be extended to the whole real axis are the linear functions and those containing $\arctan$.
Proof: A small calculation shows that the two new families are indeed solutions of the
functional equation. For the second  one, it suffices to do it for positive $b$.
Here we use an addition formula for
$\arctan$:
$$\arctan(u)+\arctan(v)=\arctan(\tfrac{u+v}{1-uv})+\pi,$$
if $u,v\geq0$ and $uv>1$. This allows to show the functional equation for sufficiently
large $x,y$; for all $x,y$ it follows because both sides of  the functional equation are real analytic functions of $x,y$.
Another calculation shows that these new families consist of reciprocally convex/concave functions
depending on the signs of $a$ and $b-c$ for the first one and of the sign of
$a$ for the second one in the case $bc\geq0.$
If $bc<0$, then $\arctan(bx+c)$ is not concave as its derivative $\dfrac b{(bx+c)^2+1}$ is not
monotonous on $(0,\infty)$.
It is much harder to show that these are all solutions. Actually, I did not guess the new family
of solutions, but found them this "hard way".
Consider a differentiable solution $f$ of the functional equation such that $f'(x)\neq0$. Then, by Darboux's Theorem,
$f'$ does not change sign, hence $f$ is strictly monotonous and its inverse function exists.
Claim 1: $f$ is infinitely often differentiable.
This follows from the functional equation which can be written
$$\frac{xf'(x)+yf'(y)}{f'(x)+f'(y)}=f^{-1}(f(x)+f(y)-f(\tfrac{x+y}2)).$$
We fix $y$. Then the right hand side is a differentiable function of $x$. We name it
$q(x)$. Then also $f'(x)=(y-q(x))f'(y)/(q(x)-x)$ is differentiable for $x\neq y$.
Here we use that $x<q(x)<y$ if $x<y$ and  $x>q(x)>y$ if $x>y$. Using several $y$, we see that
$f'(x)$ is differentiable for all $x$ and hence $f$ is twice differentiable.
This reasoning can be repeated
and yields that $f$ is infinitely often differentiable.
Claim 2: $f$ satisfies the  differential equation $\tag{1}f^{(4)}f'^2-6f^{(3)}f''f'+6f''^3=0.$
To show this, we replace $x$ by $x+h$ and $y$ by $x-h$ in the functional equation and obtain
$$f(x+h)+f(x-h)=f(x)+f\left(x+h\frac{f'(x+h)-f'(x-h)}{f'(x+h)+f'(x-h)}\right).$$
Now we write down the Taylor expansion of $f(x+h)$
$$f(x+h)=a+bh+\frac12ch^2+\frac16dh^3+\frac1{24}eh^4+O(h^5)$$
with $a=f(x)$, $b=f'(x)$, etc. and expand
both sides of the above equation up to $h^4$. This gives the differential equation.
It is not surprising to obtain a fourth order equation as a family of solutions contains up to
four parameters: With $f(x)$ also $af(bx+c)+d$ are solutions.
It was surprising for me, however, that this third order ODE for $f'$ can be explicitly
solved. In the sequel,I present my original solution. A simpler solution can be found in Edit 2 at the end of this post.
To solve (1), we first introduce $d=\dfrac{f''}{f'}$. As then
$d'=\dfrac{f^{(3)}}{f'}-\dfrac{f''^2}{f'^2}$ and $d''=
 \dfrac{f^{(4)}}{f'}-3\dfrac{f^{(3)}f''}{f'^2}+2\dfrac{f''^3}{f'^3}$, we obtain a second order
ODE for $d$:
$$\tag{2}d''-3d'd+d^3=0.$$
This equation is actually autonomous. An evident solution is $d=0$. It is the only constant
solution. This leads to the family $f(x)=ax+b$.
Now we want to find the non-constant solutions of (2).
On intervals on which $d$ is strictly monotonous, we introduce the composition
$g=d'\circ d^{-1}$ of $d'$ and the inverse function of $d$. Then
$d'(x)=g(d(x))$ and $d''(x)=g'(d(x))g(d(x))$. Replacing $d(x)$ by $s$, this gives an ODE for $g$
$$\tag{3}g'g-3sg+s^3=0.$$
It can be simplified introducing $h$ such that $g(s)=s^2h(s)$. This leads to the following ODE
for $h$ which turns out to be separable.
$$\tag{4}shh'+2h^2-3h+1=0.$$
This equation has the constant solutions $h=1$ and $h=1/2.$ They lead to
$g=s^2$ and $g=\frac12 s^2$, respectively, and thus to $d=-\frac 1{x+c}$ and $d=-\frac{2}{x+c}$
and finally to $f=a\ln(x+c)+b$ and $f=\frac a{x+c}+b$, respectively,
which are solutions already mentioned
in the question. Solving the separable equation (4) in the classical way (details omitted)
for the general solution
leads to $\frac{(h-1)^2}{2h-1}=\frac C{s^2}$ with a certain constant $C$. The equation
for $g$ is then
$\tag{5}g^2-2(s^2+C)g+s^2(s^2+C)=0.$
As its discriminant is $4C(s^2+C)$, we continue with positive $C$ and write $C=c^2$
with positive $c$. In the case of negative $C$, we could proceed differently, but
it is simpler to consider the final result with purely imaginary $c$.
So we find
$$g(s)=s^2+c^2\pm c\sqrt{s^2+c^2}.$$
Next, we need to solve the separable equation
$$\tag{6}d'=g(d)=d^2+c^2\pm c\sqrt{d^2+c^2}.$$
This is actually possible, because
$$\int\frac{ds}{s^2+ c^2\pm c\sqrt{s^2+c^2}}=\frac{-2}{s+c\pm\sqrt{s^2+c^2}}+\mbox{ constant}$$
as can be verified by differentiation.
The result is that only the negative sign is possible in (5) and that,
neglecting the constant of integration for a moment,
$$\frac{f''}{f'}=d=-\frac{2cx+2}{cx^2+2x}=-\frac c{cx+2}-\frac1{x}.$$
The general solution $d$ is obtained by replacing $x$ by $x+b$ with the
constant of integration $b$.
This last linear first order equation is solved by
$f'(x)=\dfrac{2a}{x(cx+2)}=\dfrac{a}{x}-\dfrac{ac}{cx+2}$
with another constant $a$.
This finally leads to
$$\tag{7}f(x)=a\ln\left(\frac{x+b}{c(x+b)+2}\right)+d,$$
where we reintroduced the constant of integration $b$.
Rewriting this expression a bit and renaming the parameters $a,b,c,d$
we arrive at the first new family mentioned in the beginning.
In the case of negative $C$ in (5), we just consider purely imaginary $c=i\tilde c$ in (7).
Rewriting this equation a bit and using the formula
$$\arctan(z)=\frac1{2i}\log\left(\frac{1+iz}{1-iz}\right),$$
we arrive for certain (maybe complex) values of $a,b,c,d$ in (7)
at
$$f(x)=\tilde a\arctan(\tilde c x+\tilde b)+\tilde d$$
where $\tilde a,\tilde b,\tilde c,\tilde d$ are real. This finally proves the statement
that there are only the solutions of the functional equation mentioned in the beginning
of the answer.
Edit: There is a small gap in the proof. The function $g=d'\circ d^{-1}$ is defined on $d$-images of intervals on which $d$ is strictly monotonous. It might therefore be possible that solutions $d$ or $f$ are expressed by different formulas on different subintervals. This can, however, be excluded. As a solution of the differential equation (1) and as $f'(x)\neq0$, $f$ is actually real analytic. Therefore the identity theorem for analytic functions implies that the equations $f(x)=ax+b$ or ... are valid throughout the interval of definition of $f$. This closes the gap.
Edit 2: There is a simpler way to solve the 4-th order ODE (1), $f'(x)\neq0$, but it was not so obvious.
Put $q=1/f'$. Then
$$q'=-\frac{f''}{f'^2},\ \ q''=2\frac{f''^2}{f'^3}-\frac{f^{(3)}}{f'^2},\ \ q^{(3)}=-6\frac{f''^3}{f'^4}+6\frac{f^{(3)}f''}{f'^3}-\frac{f^{(4)}}{f'^2}=0.$$
Hence $q$ is a non-zero polynomial of degree at most 2 and
$$f'(x)=\frac1{Ax^2+Bx+C},$$
where $A,B,C$ are constants and do not vanish simultaniously.
If $A=B=0,C\neq0$ then we find $f(x)=\frac1Cx+E$ with some constant $E$.
If $A=0,B\neq0$ then we find $f(x)=\frac1B\ln(x+\frac CB)+E$ and $C/B$ has to be positive since $f$ is defined on $(0,\infty)$.
If $A\neq0$ then we must distinguish 3 cases for the discriminant $D=B^2-4AC$.

*

*If $D>0$ then we can factor $Ax^2+Bx+C=A(x+b)(x+c)$ with real $b,c$. These have to be positive since $f$ is defined on $(0,\infty)$. Integrating we find
$$f(x)=a\ln\left(\frac{x+b}{x+c}\right)+E$$
with certain $a,E$.


*If $D=0$ then we can write $Ax^2+Bx+C=A(x+b)^2$ with $b=\frac B{2A}$. Again $b$ has to be positive. We obtain
$$f(x)=\frac{a}{x+b}+E.$$


*If $D<0$, $D=-F^2$ with positive $F$ then we can write
$Ax^2+Bx+C=A(x+\frac B{2A})^2+\frac{F^2}{4A}$. We obtain
$$f(x)=a\arctan(bx+c)+d$$
with certain $a,b,c,d$.
Since $f$ is real analytic as a solution of (1) with $f'(x)\neq0$, these expressions for $f$ are valid throughout $(0,\infty)$ by the identity theorem for analytic functions.
