Prove Young's inequality Suppose that f is continuous increasing function with $f(0)=0$ then prove that for $a,b>0$ we have Young's inequality 
$$ab\leq \int_{0}^{a} f(x)dx+\int_{0}^{b}f^{-1}(x)dx$$
My attempt is incorrect: I didn't argue using lower and upper sums which i assume you need to use, any help would be appreciated thanks.
 A: We can make use of the Riemann-Stieltjes integral.  Proceeding we enforce the substitution $x=f(t)$ to obtain
$$\begin{align}
\int_0^b f^{-1}(x)\,dx&=\int_{0}^{f^{-1}(b)} t\,df(t)\\\\
&=bf^{-1}(b)-\int_0^{f^{-1}(b)}f(t)\,dt\tag 1
\end{align}$$


ASIDE:
Continuous monotonic functions are differentiable almost everywhere (i.e., everywhere excluding a set of measure $0$).  But it is important to stress that in using the Riemann-Stieltjes Integral in $(1)$, we did not assume that $f$ was differentiable.


Using $(1)$ we see that 
$$\int_0^a f(x)\,dx+\int_0^b f^{-1}(x)\,dx=\int_{f^{-1}(b)}^af(x)\,dx+bf^{-1}(b)\tag 2$$

If $a>f^{-1}(b)$, then $\int_{f^{-1}(b)}^af(x)\,dx\ge b(a-f^{-1}(b))$ since $f$ is increasing.  Therefore, we assert that 
$$\int_0^a f(x)\,dx+\int_0^b f^{-1}(x)\,dx\ge ab \tag 3$$

If $a<f^{-1}(b)$, then $\int_{f^{-1}(b)}^af(x)\,dx=-\int_a^{f^{-1}(b)}f(x)\,dx\ge -b(f^{-1}(b)-a)$ since $f$ is increasing.  Therefore, we assert that 
$$\int_0^a f(x)\,dx+\int_0^b f^{-1}(x)\,dx\ge ab \tag 4$$

Finally, if $a=f^{-1}(b)$, then we see directly from $(2)$ that
$$\int_0^a f(x)\,dx+\int_0^b f^{-1}(x)\,dx\ge ab \tag 5$$

Putting together $(3)$, $(4)$, and $(5)$ yields the coveted result
$$\bbox[5px,border:2px solid #C0A000]{\int_0^a f(x)\,dx+\int_0^b f^{-1}(x)\,dx\ge ab}$$
as was to be shown!
A: Not exactly an answer, but too long to comment.
There exists a formula for the antiderivative of an inverse function. If we let $f(x)$ have inverse $f^{-1}(x)$ and antiderivative $F(x)$, then
$$xf^{-1}(x)-(F\circ f^{-1})(x)+C$$
and if we plug this in, we get
$$F(a)-F(0)+bf^{-1}(b)-(F\circ f^{-1})(b)-0f^{-1}(0)+(F\circ f^{-1})(0)$$
$$F(a)-F(0)+bf^{-1}(b)-(F\circ f^{-1})(b)+(F\circ f^{-1})(0)$$
If $f(x)$ is continuously increasing then it must be one-to-one and it has an inverse, so $f^{-1}(0)=0$ and we get
$$F(a)-F(0)+bf^{-1}(b)-(F\circ f^{-1})(b)+F(0)$$
$$F(a)+bf^{-1}(b)-(F\circ f^{-1})(b)$$
Does this simplify it enough for you to take it from here?
