Showing that $f(x)=\frac x{x+1}$ is the unique function satisfying $f(x)+f\left(\frac1x\right)=1$ and $f(2x)=2f\big(f(x)\big)$ We are given a function $ f : \mathbb Q ^ + \to \mathbb Q ^ + $ such that $$ f ( x ) + f \left( \frac 1 x \right) = 1 $$
and
$$ f ( 2 x ) = 2 f \big( f ( x ) \big) \text . $$
Find, with proof, an explicit expression for $f(x)$ for all positive rational numbers $x$.
Every number I have evaluated is of the form $ f ( x ) = \frac x { x + 1 } $ and this clearly fits the functional equations, but I can't prove that it's the only solution. Can anyone help me? I have put down the start of my workings which led me to the conjecture of $ f ( x ) = \frac x { x + 1 } $.
Plugging in $ x = 1 $ clearly gives $ f ( 1 ) = \frac 1 2 $ and $ f ( 2 ) = 2 f \big( f ( 1 ) \big) = 2 f \left( \frac 1 2 \right) $ which we can plug back into the first equation to get that $ f ( 2 ) = \frac 2 3 $. Working in this vein I have been able to show that $ f ( x ) = \frac x { x + 1 } $ for particular values of $ x $, but not in general.
The most difficult part appears to be proving it for the even integers. To prove $ x = 8 $, we have
$$ f ( 12 ) = 2 f \left( \frac 6 7 \right) = 4 f \left( \frac 3 { 10 } \right) = 4 - 4 f \left( \frac { 10 } 3 \right) \\
= 4 - 8 f \left( \frac 5 8 \right) = 8 f \left( \frac 8 5 \right) - 4 = 16 f \left( \frac 4 9 \right) - 4 \\
= 32 f \left( \frac 2 { 11 } \right) - 4 = 64 f \left( \frac 1 { 12 } \right) - 4 = 60 - 64 f ( 12 ) \text , $$
giving us $ f ( 12 ) = \frac { 12 } { 13 } $. This will probably be the main area of difficulty in the proof.

For each coprime $n, m \in \mathbb N_+$, $f\left( \dfrac nm\right)=\dfrac{n}{n+m}$ is a solution.
To prove this, we induct on $n+m$. The base case is clearly true. Suppose it's true for $n+m\leq k$ for some natural number $k$ and consider $n+m=k+1$.
We consider two cases:
Case 1, $n>m$:
Using the first condition and Case 2, we get that
$$f\left(\dfrac{n}{m}\right) = 1 - f\left(\dfrac{m}{n} \right)=1-\frac{m}{m+n}=\frac{n}{n+m}\,\,\,\,\,\,\,\,\,\blacksquare$$
Case 2, $m>n$:
By the condition, we can assume $m=n+l$ for some natural $l$. Then,
\begin{align*}f\left( \dfrac{n}{m}\right)=f\left( \frac{n}{n+l}\right) = f\left( f\left( \frac{n}{l}\right)\right) = \frac 12f\left( \frac{2n}{l}\right) = \frac{n}{2n+l} = \frac{n}{n+m}\,\,\,\,\,\,\,\,\,\,\blacksquare
\end{align*}
Finally, we plug the function in and see that it indeed satisfies the conditions.
This was problem B4 of Ireland MO 1991.
