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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.

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    look at legendre transforms too2012-05-26

2 Answers 2

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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.

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    @Frank - I will second Marvis' comment - I've only used Young's Inequality to prove the Triangle Inequality in Lp spaces as well. Such a proof seems to be a canonical introduction to the inequality in a course in real analysis (at least, that's where I first encountered it).2012-09-18
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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.