$\int_{0}^{a} f(t) dt +\int_{0}^{b} f^{-1}(t)dt \geq ab$ Want to prove
$$\int_{0}^{a} f(t) dt +\int_{0}^{b} f^{-1}(t)dt \geq ab$$
where $f$ is continuous, strictly increasing and $f(0)=0$, $a>0$ and $b\in f((0,\infty))$.
I know how to do the question if $f$ is differentiable, which uses integration by substitution. But when the condition is changed to continuous, I cannot even think of a direction.
Thank you.

Substituting $t\mapsto f(t)$ yields
$$
\begin{align}
\int_0^bf^{-1}(t)\,\mathrm{d}t
&=\int_0^{f^{-1}(b)}t\,\mathrm{d}f(t)\\
&=bf^{-1}(b)-\int_0^{f^{-1}(b)}f(t)\,\mathrm{d}t\tag{1}
\end{align}
$$
If $a\ge f^{-1}(b)$, then
$$
\begin{align}
\int_0^af(t)\,\mathrm{d}t+\int_0^bf^{-1}(t)\,\mathrm{d}t
&=bf^{-1}(b)+\int_{f^{-1}(b)}^af(t)\,\mathrm{d}t\\
&\ge bf^{-1}(b)+(a-f^{-1}(b))b\\[9pt]
&=ab\tag{2}
\end{align}
$$
Substituting $t\mapsto f^{-1}(t)$ yields
$$
\begin{align}
\int_0^af(t)\,\mathrm{d}t
&=\int_0^{f(a)}t\,\mathrm{d}f^{-1}(t)\\
&=af(a)-\int_0^{f(a)}f^{-1}(t)\,\mathrm{d}t\tag{3}
\end{align}
$$
If $f(a)\lt b$, then
$$
\begin{align}
\int_0^af(t)\,\mathrm{d}t+\int_0^bf^{-1}(t)\,\mathrm{d}t
&=af(a)+\int_{f(a)}^bf^{-1}(t)\,\mathrm{d}t\\
&\ge af(a)+(b-f(a))a\\[9pt]
&=ab\tag{4}
\end{align}
$$
