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.


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. enter image description here

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


For positives $a$, $b$, $p$ and $q$ we'll rewrite our inequality in the following form. $$\ln\left(\frac{a^p}{p}+\frac{b^q}{q}\right)\geq\frac{1}{p}\ln{a^p}+\frac{1}{q}\ln{a^q},$$ which is just Jensen for the concave function $\ln$, which is geometry, of course.