Spivak, Chapter 13, "Integrals" Problem 22: Prove Young's Inequality The following is a problem from chapter 13, "Integrals", from Spivak's Calculus



*Suppose that $f$ is a continuous increasing function with $f(0)=0$. Prove that for $a,b>0$ we have Young's Inequality,

$$ab\leq \int_0^a f(x)dx+\int_0^b f^{-1}(x)dx\tag{1}$$
and that equality holds if and only if $b=f(a)$. Hint: Draw a picture
like Figure 16.

The cited picture 16 isn't that important, it is just an example

When one draws a picture, $(1)$ is intuitively pretty obvious

Above is the case where $b>f(a)$.
I spent a while thinking about how to prove $(1)$ analytically. Finally, I looked at the solution manual and the solution there was just the obvious intuitive solution I had first thought of, but without the equations I spent time trying to find. Here it is

The figure below shows the case $b<f(a)$.

We have $$ab=\text{ area } OACB<\text{ area } OAE+ \text{ area } OBD$$
$$=\int_0^a f(x)dx+\int_0^bf^{-1}(x)dx$$
If $b=f(a)$ we clearly have equality. It is easy to see that we have
the same inequality if $b>f(a)$ (or simply apply the first inequality
to $f^{-1}$).

Is this proof considered rigorous? Is there one that uses equations instead of words like "clearly".
 A: Assume $b>f(a)$.
Because $f$ is increasing, so is $f^{-1}$, and therefore for any $x$ larger than $f(a)$, we have $f^{-1}(x)>f^{-1}(f(a))=a$
$$f^{-1}(f(a))(b-f(a))=a(b-f(a))<\int_{f(a)}^b f^{-1}(x)dx\tag{1}$$
$$=\int_0^{b}f^{-1}(x)dx-\int_0^{f(a)} f^{-1}(x)dx\tag{2}$$
$$ab<af(a)+\int_0^{b}f^{-1}(x)dx-\int_0^{f(a)} f^{-1}(x)dx\tag{3}$$
In problem 21 of this same chapter, Spivak asks us to prove
$$\int_a^b f^{-1}=bf^{-1}(b)-af^{-1}(a)-\int_{f^{-1}(a)}^{f^{-1}(b)}f\tag{4}$$
which is equivalent to
$$\int_a^b f=bf(b)-af(a)-\int_{f(a)}^{f(b)}f^{-1}\tag{5}$$
and if we take $a=0$ and $f(a)=0$ in $(5)$ we get
$$\int_0^b f=bf(b)-\int_0^{f(b)}f^{-1}\tag{6}$$
But we can see the right-hand terms of $(6)$ on the right-hand side of $(3)$. Thus from $(3)$ we have
$$ab<\int_0^a f(x)dx+\int_0^{b}f^{-1}(x)dx\tag{7}$$
Which is what we wanted to prove.
The $b<f(a)$ case is analogous and symmetric to the case above.
Finally, for the $b=f(a)$ case, if we start from $6$ we have
$$\int_0^a f = af(a)-\int_0^{f(a)}f^{-1}=ab-\int_0^{b}f^{-1}$$
$$\implies ab = \int_0^a f+\int_0^{b}f^{-1}$$
