Show that $\int_{0}^{a}f(x)\mathrm{d}x + \int_{0}^{b}f^{-1}(y)\mathrm{d}y \geq ab$

Question

Let $f:[0,\infty[\to\mathbb{R}$ be a continuous and strictly increasing function with $f(0) = 0$. Let $a$ and $b$ be two positive arbitrary numbers. Show that $$ \int_{0}^{a}f(x)\mathrm{d}x + \int_{0}^{b}f^{-1}(y)\mathrm{d}y \geq ab \quad. $$

I tried doing integration by parts on the first integral and then making the substitution $y = f(x)$ and I got $$ \int_{0}^{a}f(x)\mathrm{d}x = af(a) - \int_{0}^{f(a)}f^{-1}(y)\mathrm{d}y $$ and then substituting it on the inequality $$ af(a) + \int_{0}^{b}f^{-1}(y)\mathrm{d}y \geq ab + \int_{0}^{f(a)}f^{-1}(y)\mathrm{d}y $$ but I have no clue on where to go from here

Answer 1

If you draw a picture, it is very easy to see why this is true.

Define $c=f^{-1}(b)\Leftrightarrow f(c)=b$

If $c\le a$

$$\begin{align}ab&=\int_0^b f^{-1}(y)dy+\int_0^c f(x)dx+b(a-c)\\ \\ &\le\int_0^b f^{-1}(y)dy+\int_0^c f(x)dx+\int_c^a f(x)dx\\ \\ &=\int_0^b f^{-1}(y)dy+\int_0^a f(x)dx\end{align}$$

and if $c>a$, then

$$\begin{align}ab&=\int_0^a f(x)dx+\int^{f(a)}_0 f^{-1}(y)dy+a(b-f(a))\\ \\ &\le \int_0^a f(x)dx+\int^{f(a)}_0 f^{-1}(y)dy+\int_{f(a)}^b f^{-1}(y)dy\\ \\ &=\int_0^a f(x)dx+\int_0^b f^{-1}(y)dy \end{align}$$