## Spring 2018, problem 60

Let $f,g:[a,b]\to [0,\infty)$ be two continuous and non-decreasing functions such that each $x\in [a,b]$ we have $$\int^x_a \sqrt { f(t) }\ dt \leq \int^x_a \sqrt { g(t) }\ dt \ \ \textrm{and}\ \int^b_a \sqrt {f(t)}\ dt = \int^b_a \sqrt { g(t)}\ dt.$$ Prove that $$\int^b_a \sqrt { 1+ f(t) }\ dt \geq \int^b_a \sqrt { 1 + g(t) }\ dt.$$

Let us define the functions $\varphi,\gamma:[a,b]\to[0,\infty)$ by means of: $$\varphi(x):=\int_a^x\sqrt{f(t)}\,dt\,;\quad\gamma(x):=\int_a^x\sqrt{g(t)}\,dt$$ so that the functions $\varphi,\gamma$ are convex and satisfy: $$\varphi(x)\le\gamma(x)\text{ on }[a,b]\,;\ \varphi(a)=\gamma(a),\ \varphi(b)=\gamma(b)\,.$$ Denoting by $G_\varphi,\,G_\gamma$ the graphs of the functions, the thesis can be reformulated as: $$Length\left(G_\varphi\right)\,\ge\,Length\left(G_\gamma\right)$$ and the result is a well known result in Calculus of Variations (the convexity of $\varphi$ being not needed): for a convex "obstacle" $\gamma$, among all the curves remaining below $\gamma$ and having equal values at the extremes, $\gamma$ itself is the one of minimal length.