XXXI Brazilian Math Olympiad

First notice that, since $f$ is increasing and has an inverse, then it is bijective and hence $f(0)=0$ and $f(1)=1$ (if $f(0)>0$ then $f$ would never be equal to $0$; the same applies if $f(1)<1$). So, by applying the substitution $x=f(y)$, for which $\frac{dx}{dy}=f^{\prime }(y)$, and integration by parts, $$\int f^{-1}(x)dx=\int f^{-1}(f(y))f^{\prime }(y)dy=\int y{f}^{\prime }(y)dy=yf(y)-\int f(y)dy$$ Thus $$\begin{gathered} {\int }_0^1{f}^{-1}(x)dx={yf(y)|}_0^1-{\int }_0^{1}f(y)dy=f(1)-0-{\int }_0^{1}f(x)dx \\ ⟺{\int }_0^{1}f(x)dx+{\int }_0^1{f}^{-1}(x)dx=1 \end{gathered}$$ and both definite integrals are equal to $\frac{1}{2}$. Since ${\int }_0^{1}xdx=\frac{1}{2}$ as well, $${\int }_0^1(f(x)-x)dx=0$$ Let $g(x)=f(x)-x$. Then $g(0)=f(0)-0=0$ and $g(1)=f(1)-1=0$. Since ${\int }_0^{1}g(x)dx=0$, there exist two real numbers $0<t<u<1$ such that $g(t)\le 0$ and $g(u)\ge 0$, and so there is $c\in ]t,u[$ such that $g(c)=0$. Then, by the mean value theorem (or Rolle's theorem) there exist $a\in ]0,c[$ and $b\in ]c,1[$ such that $g^{\prime }(a)=g^{\prime }(b)=0$. Since $g(x)=f(x)-x⟹g^{\prime }(x)=f^{\prime }(x)-1$ and so $f^{\prime }(a)=f^{\prime }(b)=1$.