calculusinequalityintuitionvisualizationyoung-inequality
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.
The image was made using grapher and some post processing was done using LaTeXiT and preview on Mac OSX.
You can prove it by contradiction. Suppose that $u^p>v^q$.
Let $f(x)=x^{p-1}$. Then $u^p>v^q\Leftrightarrow u^p>v^{p(q-1)}\Leftrightarrow u^{\frac{1}{q-1}}>v \Leftrightarrow u^{p-1}>v \Leftrightarrow f(u)>v$. Let's take a look at rectangle with sides equals to $u$ and $v$. Then using usual geometric proof of Young's inequality we see that $$\dfrac{u^p}{p}+\dfrac{v^q}{q}>uv.$$ Analogous reasoning for $u^p<v^q$.
Best Answer
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.