THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

Obviously $f$ is increasing. Using Lagrange's theorem on $[f(a),f(b)]$ we get a point $c_1\in (f(a),f(b))$ such that $$f(f(b))-f(f(a))=f^{\prime }(c_1)(f(b)-f(a)).$$ Another use of the theorem for the interval $[a,b]$ gives a point $c_2\in (a,b)$ such that $$f(b)-f(a)=f^{\prime }(c_2)(b-a).$$ Collecting, we get $$f(f(b))-f(f(a))=f^{\prime }(c_1)f^{\prime }(c_2)(b-a).$$ As $f^{\prime }>0$ the number $\sqrt{f^{\prime }(c_1)f^{\prime }(c_2)}$ is between $f^{\prime }(c_1)$ and $f^{\prime }(c_2)$. Because $f^{\prime }$ has the intermediate point property (Darboux's theorem), there is a point $c$, between $c_1$ and $c_2$ such that $f^{\prime }(c)=\sqrt{f^{\prime }(c_1)f^{\prime }(c_2)}$. This gives $$f^{\prime }(c_1)f^{\prime }(c_2)=f^{\prime }(c{)}^2,$$ and, in conclusion, $$f(f(b))-f(f(a))=f^{\prime }(c{)}^2(b-a),c\in (a,b).$$