Solution-4

Again, let us prove that \(f(x)=1 / x\) is the only solution. Let again \(g(x)\) be the unique positive real number such that \((x, g(x))\) is a good pair.

Lemma 4: The function \(f\) is strictly convex.

Proof: Consider the function \(q_{s}(x)=f(x)+s x\) for some real number \(s\). If \(f\) is not strictly convex, then there exist \(u<v\) and \(t \in(0,1)\) such that

\[f(t u+(1-t) v) \geqslant t f(u)+(1-t) f(v)\]

Hence

\[\begin{aligned} q_{s}(t u+(1-t) v) & \geqslant t f(u)+(1-t) f(v)+s(t u+(1-t) v) \\ & =t q_{s}(u)+(1-t) q_{s}(v) \end{aligned}\]

Let \(w=t u+(1-t) v\) and consider the case \(s=f(g(w)) / g(w)\). For that particular choice of \(s\), the function \(q_{s}(x)\) has a unique minimum at \(x=w\). However, since \(q_{s}(w) \geqslant t q_{s}(u)+(1-t) q_{s}(v)\), it must hold \(q_{s}(u) \leqslant q_{s}(w)\) or \(q_{s}(v) \leqslant q_{s}(w)\), a contradiction. **Lemma 5:** The function \(f\) is continuous. *Proof:* Since \(f\) is strictly convex and defined on an open interval, it is also continuous. As in Solution 1, we can prove that \(f(x) \leqslant 1 / x\). If \(f(x)<1 / x\), then we consider the function \(h(y)=x f(y)+y f(x)\) which is continuous. Since \(h(x)<2\), there exist at least two distinct \(z \neq x\) such that \(h(z)<2\) giving that \((x, z)\) is good pair for both values of \(z\), a contradiction. We conclude that \(f(x)=1/x\) as desired.

Comment: Lemma 5 implies Lemma 3, using an argument similar as in the end of Solution 4.