Prove that for $a,b > 0$, we have $\int_{0}^a f(x)dx+\int_{0}^b f^{-1}(x)dx \geq ab$ 
Suppose $f$ is a continuous, strictly increasing function over $\mathbb{R}$ with $f(0) = 0$.  Prove that for $a,b > 0$, we have $\displaystyle \int_{0}^a f(x)dx+\int_{0}^b f^{-1}(x)dx \geq ab$. When does the equality hold? 

Attempt 
It is easy to see that for $a \neq c$ we have $$\displaystyle \int_{0}^a f(t) dt > \int_{0}^c f(t)dt + f(c)(a-c)$$ with equality if and only if $a = c$. Therefore, if $c = f^{-1}(b)$ then $$\displaystyle \int_{0}^a f(t)dt \geq \int_{0}^{f^{-1}(b)}f(t)dt + ab -bf^{-1}(b).$$ What would be the easiest way to finish from here?
 A: Let's begin with the approach in the OP.  Then, we have
$$\int_0^a f(x)\,dx\ge \int_0^{f^{(-1)}(b)} f(t)\,dt + ab - bf^{(-1)}(b) \tag 1$$
Since the integral on the right-hand side of $(1)$ can be written as
$$\int_0^{f^{(-1)}(b)} f(t)\,dt=bf^{(-1)}(b)-\int_0^b f^{(-1)}(x)\,dx$$
we find after simplifying and rearranging terms the coveted inequality
$$\int_0^a f(x)\,dx + \int_0^{b} f^{-1}(x)\,dx \ge  ab$$ 
with equality if and only if $f(a)=b$.  And we are done!

Alternative Approach
Here, we present an alternative approach under the additional assumption that $f(x)$ is differentiable.  To that end, we proceed.
Let $G(a,b)$ be the function given by 
$$G(a,b)=\int_0^a f(x)\,dx+\int_0^b f^{(-1)}(x)\,dx-ab$$
Note that $G(0,0)=0$.  Now, note that we have the first partial derivatives
$$\begin{align}
\frac{\partial G(a,b)}{\partial a}&=f(a)-b\\\\
\frac{\partial G(a,b)}{\partial b}&=f^{(-1)}(b)-a
\end{align}$$
and the second partial derivatives
$$\begin{align}
\frac{\partial^2 G(a,b)}{\partial a^2}&=f'(a)\\\\
\frac{\partial^2 G(a,b)}{\partial b^2}&=\left(f^{(-1)}\right)'(b)\\\\
\frac{\partial^2 G(a,b)}{\partial a \partial b}&=-1\\\\
\end{align}$$
Since $f$ is strictly monotonically increasing, $f(a)\left(f^{(-1)}\right)'(b)>0$, and the determinant of the Hessian is, therefore, always positive.  Hence, $G(a,b)$ is a minimum when $f(a)=b$ and $f^{(-1)}(b)=a$.  The minimum is therefore,
$$\min_{(a,b)}G(a,b)=\int_0^a f(x)\,dx+\int_0^{f(a)} f^{(-1)}(x)\,dx-ab=0$$
Therefore, $G(a,b)\ge 0$ with the equality holding only for $(a,b)=(a,f(a))$.  Finally, we can write for $a>0$ and $b>0$
$$\int_0^a f(x)\,dx+\int_0^b f^{(-1)}(x)\,dx\ge ab$$
with equality only when $b=f(a)$. 
A: You can imagine your curve as a squiggle in the first quadrant that moves up and to the right (starting at the origin).  Draw any shape like this (even a straight line).  The first integral
$$
\int_0^af(x)dx
$$
calculates the area in the region below the curve for $x$-values less than $a$.  Shade this region in your drawing (it is like a triangular region under the curve whose base is the $x$-axis).
The second integral
$$
\int_0^bf^{-1}(x)dx
$$
calculates the area to the left of the curve and below the line $y=b$.  Shade this region in your drawing (it is like a triangular region to the left of the curve whose base is the $y$-axis).
Consider the rectangle whose dimensions are $a$ on the $x$-axis and $b$ on the $y$ axis.  This rectangle has area $ab$ and is within your shaded region, and, this gives your inequality (the area on the LHS contains the area on the RHS).
The case where they're equal is where the upper right corner of the rectangle is on the curve and is the endpoint of integration (in other words, when the region of integration is exactly the rectangle).  This occurs only when $f(a)=b$.
