Geometric interpretation of Young's inequality

Question

Is there a geometric interpretation of Young's inequality, $$ab \leq \frac{a^{p}}{p} + \frac{b^{q}}{q}$$ with $\dfrac{1}{p}+\dfrac{1}{q} = 1$?

My attempt is to say that $ab$ could be the surface of a rectangle, and that we could also say that:

$\dfrac{a^{p}}{p}=\displaystyle \int_{0}^{a}x^{p-1}dx$,

but them I'm stuck.

Answer 1

First note that we have $$ab \leq \int_0^{a} f(x) dx + \int_0^{b} f^{-1}(x) dx $$ for any strictly increasing integrable function $f(x)$. The geometric interpretation is from looking at the area of the rectangle with coordinates $(0,0)$,$(a,0)$,$(a,b)$ and $(0,b)$ and comparing it with the areas given by the integrals. From the image it is also clear that the equality hold only when $b=f(a)$.

To get the Young's inequality, choose $f(x) = x^{p-1}$.

I have added the following picture for clarity.

The image was made using grapher and some post processing was done using LaTeXiT and preview on Mac OSX.