$\int_{0}^{a} f(t) dt +\int_{0}^{b} f^{-1}(t)dt \geq ab$ 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.
 A: 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
$$
\begin{align}
\int_0^af(t)\,\mathrm{d}t+\int_0^bf^{-1}(t)\,\mathrm{d}t
&=af(a)+\int_{f(a)}^bf^{-1}(t)\,\mathrm{d}t\\
&\ge af(a)+(b-f(a))a\\[9pt]
&=ab\tag{4}
\end{align}
$$
A: From the properties of $f$ and $f^{-1}$, both strictly increasing continuous functions, the following properties hold:
The Fubini-Tonelli Theorem justifies
$$
\int_{0}^{a} f(t) dt +\int_{0}^{f(a)} f^{-1}(t)dt = \int_{0}^{a} \int_{0}^{f(t)}  \text ds\text dt +\int_{0}^{f(a)} \int_{0}^{f^{-1}(t)} \text ds\text dt \\= \int_{0}^{f(a)}\int_{f^{-1}(s)}^{a}\text dt\text ds + \int_{0}^{f(a)}\int^{f^{-1}(s)}_{0}\text dt\text ds = \int_{0}^{f(a)}\Big(\int_{f^{-1}(s)}^{a} \text dt + \int^{f^{-1}(s)}_{0}\text dt\Big)\text ds \\= \int_{0}^{f(a)}\Big(\int_{0}^{a} \text dt\Big)\text ds = af(a)\,.
$$
That is,

$$\int_{0}^{a} f(t) dt +\int_{0}^{f(a)} f^{-1}(t)dt = af(a)\tag{1}$$

A similar argument shows

$$\int_{0}^{f^{-1}(b)} f(t)dt + \int_{0}^{b} f^{-1}(t)dt =  bf^{-1}(b)\tag{2}$$

Furthermore,

$$
\int_{f(a)}^{b} f^{-1}(t)dt\ \geqslant\ f(f^{-1}(a))\int^{b}_{f(a)}\text dt\ =\ a(b-f(a))\tag{3}
$$

And, finally,

$$
\int_{f^{-1}(b)}^{a} f(t) \text dt\ \geqslant\ f(f^{-1}(b))\int_{f^{-1}(b)}^{a} \text dt\ =\ b(a-f^{-1}(b))\tag{4}
$$

Using (1) to (4) we proceed as follows. 
Let $a<f^{-1}(b)$. Then
$$
\int_{0}^{a} f(t) dt +\int_{0}^{b} f^{-1}(t)dt = \int_{0}^{a} f(t) dt +\int_{0}^{f(a)} f^{-1}(t)dt + \int_{f(a)}^{b} f^{-1}(t)dt \\ \geqslant af(a) + a(b-f(a)) = ab
$$
Let $a>f^{-1}(b)$. Then
$$
\int_{0}^{a} f(t) dt +\int_{0}^{b} f^{-1}(t)dt = \int_{f^{-1}(b)}^{a} f(t) dt +\int_{0}^{f^{-1}(b)} f(t)dt + \int_{0}^{b} f^{-1}(t)dt \\ \geqslant bf^{-1}(b) + b(a-f^{-1}(b)) = ab 
$$
A: Let $C$ be the portion of the curve $y = f(x)$ where $0 \le x \le a$. Since $f$ is strictly increasing, so is $f^{-1}$. Therefore, if $f(a) \le b$, then
\begin{align}\int_0^a f(t)\, dt + \int_0^b f^{-1}(t)\, dt &= \int_0^a f(t)\, dt + \int_0^{f(a)} f^{-1}(t)\, dt + \int_{f(a)}^b f^{-1}(t)\, dt\\
& = \int_C y\, dx + x\, dy + \int_{f(a)}^b f^{-1}(t)\, dt\\
& = \int_C d(xy) + \int_{f(a)}^b f^{-1}(t)\, dt\\
& = af(a) + \int_{f(a)}^b f^{-1}(t)\, dt\\
& \ge af(a) + f^{-1}(f(a))(b - f(a))\\
& = ab.
\end{align}
A similar argument holds when $f(a) \ge b$. If $C'$ is the portion of the curve $x = f^{-1}(y)$ where $0 \le y \le b$, then
\begin{align}\int_0^a f(t)\, dt + \int_0^b f^{-1}(t)\, dt &= \int_{C'} y\, dx + x\, dy + \int_{f^{-1}(b)}^a f(t)\, dt \\
&= bf^{-1}(b) + \int_{f^{-1}(b)}^a f(t)\, dt\\
&\ge bf^{-1}(b)  + b(a - f^{-1}(b)) \\
&= ab.
\end{align}
