Problem source

The function $\mathrm{g}$ satisfies, for all positive $x$ and $y$,
		\[
		\mathrm{g}(x)+\mathrm{g}(y)=\mathrm{g}(z),\tag{*}
		\]
		where $z=xy/(x+y+1).$ By treating $y$ as a constant, show that 
		\[
		\mathrm{g}'(x)=\frac{y^{2}+y}{(x+y+1)^{2}}\mathrm{g}'(z)=\frac{z(z+1)}{x(x+1)}\mathrm{g}'(z),
		\]
		and deduce that $2\mathrm{g}'(1)=(u^{2}+u)\mathrm{g}'(u)$ for all $u$ satisfying $0 < u < 1.$ Now by treating $u$ as a variable, show that 
		\[
		\mathrm{g}(u)=A\ln\left(\frac{u}{u+1}\right)+B,
		\]
		where $A$ and $B$ are constants. Verify that $\mathrm{g}$ satisfies $(*)$ for a suitable value of $B$. Can $A$ be determined from $(*)$?
 
The function $\mathrm{f}$ satisfies, for all positive $x$ and $y$,
		\[
		\mathrm{f}(x)+\mathrm{f}(y)=\mathrm{f}(z)
		\]
		where $z=xy.$ Show that $\mathrm{f}(x)=C\ln x$ where $C$ is a constant.

Solution source

Note that $z = xy/(x+y+1) \Rightarrow y(x-z) = z(x+1)$

\begin{align*}
&& g(x) + g(y) &= g(z) \\
\Rightarrow && g'(x) &= g'(z) \cdot \frac{y(x+y+1) - xy \cdot 1} {(x+y+1)^2} \\
&&&= g'(z) \frac{y^2+y}{(x+y+1)^2} \\
&&&= g'(z) \frac{z^2(y^2+y)}{x^2y^2} \\
&&&= g'(z) \frac{z^2(y+1)}{x^2y} \\
&&&= g'(z) \frac{z^2}{x^2} \left (1 + \frac{x-z}{z(x+1)} \right) \\
&&&= g'(z) \frac{z}{x^2}  \frac{zx+x}{x+1} \\
&&&= g'(z) \frac{z(z+1)}{x(x+1)}
\end{align*}

If $x = 1$ then as $y$ ranges from $0$ to $\infty$, $z$ ranges from $0$ to $1$, so $g'(1) = \frac{z(z+1)}{1(1+1)}g'(z)$, ie $2g'(1) = (u^2+u)g'(u)$.

\begin{align*}
&& g'(u) &= \frac{A}{u(u+1)} \\
\Rightarrow && g(u) &= A\int \left ( \frac{1}{u} - \frac{1}{u+1} \right) \d u \\
&&&= A \left ( \ln u - \ln(u+1) \right) + B \\
&&&= A \ln \left ( \frac{u}{u+1} \right) + B \\
\\
&& A \ln \left ( \frac{x}{x+1} \right) + B+A \ln \left ( \frac{y}{y+1} \right) + B &=A \ln \left ( \frac{z}{z+1} \right) + B \\
\Rightarrow && B &= A \ln \left ( \frac{z}{z+1} \frac{y+1}{y} \frac{x+1}{x} \right) \\
&&&= A \ln \left ( \frac{1}{1+\frac{x+y+1}{xy}} \frac{(y+1)(x+1)}{xy} \right) \\
&&&= A \ln 1 \\
&&& = 0
\end{align*}

Therefore $B = 0$. $A$ cannot be determined from $(*)$.

Suppose $f(x) + f(y) = f(z)$, then $f'(x) = yf'(z)$. Letting $x = 1$ we find $f'(1) = uf'(u) \Rightarrow f(u) = C \ln u  + D$, but $D = 0$ so $f(x) = C \ln x$