Proof of Young's inequality The following problem is from Spivak's Calculus.
Suppose that $f$ is a continuous increasing function with $f(0)=0$. Prove that for $a,b \gt 0$ we have Young's inequality 
$$ ab \le \int_0^af(x)dx+\int_0^bf^{-1}(x)dx$$, and that equality holds if and only if $b=f(a)$.
It is enough to consider the case $f(a) \gt b$, and show that the strict inequality occurs in this case.
I've tried proving this using the theorem 
$$ \int_a^bf^{-1}=bf^{-1}(b)-af^{-1}(a)-\int_{f^{-1}(a)}^{f^{-1}(b)}f$$
but I got stuck along the way. 
How may I show this rigorously using the definition or properties of integrals? Any hint, suggestions or solutions would be appreciated.
 A: Assuming $f(a)>b$, we have:
$$ \color{red}{\int_{0}^{a} f(x)\,dx} = \mu\left(\{(x,y)\in[0,a]\times[0,f(a)]: 0\leq y\leq f(x)\}\right) $$
and:
$$ \color{blue}{\int_{0}^{b} f^{-1}(y)\,dy} = \mu\left(\{(x,y)\in[0,a]\times[0,b]: 0\leq x\leq f^{-1}(y)\}\right) $$
so the sum of the two integrals surely exceeds $\mu\left([0,a]\times[0,b]\right)=ab$. 
By the way, it is a lot easier just to draw a picture:
$\hspace3in$
A: Let $C$ be the graph of $v = f(u)$ over the interval $[0,f^{-1}(b)]$. If $f(a) > b$, then $f^{-1}(b) < a$, in which case
\begin{align}\int_0^a f(x)\, dx + \int_0^b f^{-1}(x)\, dx &= \int_0^{f^{-1}(b)} f(x)\, dx + \int_0^b f^{-1}(x)\, dx + \int_{f^{-1}(b)}^a f(x)\, dx\\
&= \int_C u\, dv + v\, du + \int_{f^{-1}(b)}^a f(x)\, dx\\
&= \int_C d(uv) + \int_{f^{-1}(b)}^a f(x)\, dx\\
&= bf^{-1}(b) + \int_{f^{-1}(b)}^a f(x)\, dx\\
&> bf^{-1}(b) + b(a - f^{-1}(b))\\
&= ab
\end{align}
Similarly if $f(a) < b$, then $\int_0^a f(x)\, dx + \int_0^b f(x)\, dx > ab$. If $f(a) = b$, then $$\int_0^a f(x)\, dx + \int_0^b f^{-1}(x)\, dx = \int_C u\, dv + v\, du = \int_C d(uv) = af(a) = ab.$$
