Show $\int_0^a f(x)\,dx + \int_0^b f^{-1}(x)\,dx \ge ab$ for strictly increasing function $f(x)$ Let $f: [0, \infty) \to [0, \infty)$ be continuous and strictly increasing and with $f(0) = 0$. Prove that $$\int_0^a f(x)\,dx + \int_0^b f^{-1}(x)\,dx \ge ab$$ for any $a, b > 0$, and give a condition for equality to hold.

Note that 
$$\int_0^cf(x)\,dx+\int_0^{f(c)}f^{-1}(x)\,dx=cf(c)$$
You can think of this as the area of the rectangle with $(0,0)$ for one corner and $(c,f(c))$ for the opposite corner.
Without loss of generality, let $f(a)\le b$. Then
$$\int_0^af(x)\,dx+\int_0^bf^{-1}(x)\,dx=af(a)+\int_{f(a)}^bf^{-1}(x)\,dx$$
Now since $f(x)$ is strictly increasing, we have $f^{-1}(x)>a$ for $x>f(a)$. Hence
$$\int_{f(a)}^bf^{-1}(x)\,dx\ge a\left(b-f(a)\right)$$
with equality if and only if $b=f(a)$.
Putting this together, we get
$$\int_0^af(x)\,dx+\int_0^bf^{-1}(x)\,dx\ge af(a)+a(b-f(a))=ab$$
with equality if and only if $b=f(a)$.
