IMC 2004 · Day 2 · P8

Let $F(x) = \int_a^x \sqrt{f(t)}\,dt$ and $G(x) = \int_a^x \sqrt{g(t)}\,dt$. The functions $F, G$ are convex, $F(a) = 0 = G(a)$ and $F(b) = G(b)$ by the hypothesis. We are supposed to show that $$\int_a^b \sqrt{1 + \bigl( F'(t) \bigr)^2}\,dt \ge \int_a^b \sqrt{1 + \bigl( G'(t) \bigr)^2}\,dt$$ i.e. The length ot the graph of $F$ is $\ge$ the length of the graph of $G$. This is clear since both functions are convex, their graphs have common ends and the graph of $F$ is below the graph of $G$ — the length of the graph of $F$ is the least upper bound of the lengths of the graphs of piecewise linear functions whose values at the points of non-differentiability coincide with the values of $F$, if a convex polygon $P_1$ is contained in a polygon $P_2$ then the perimeter of $P_1$ is $\le$ the perimeter of $P_2$.