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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.

Martin Sleziak
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Frank
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2 Answers2

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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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    Cool! That's easy to see. It's actually better in the first form you wrote down, more general it seems. Can you point out how this is useful in maths? I.e. do we use that to get an upper or lower bound? To replace sums by products or vice-versa? Or something else? Thanks in any case! – Frank May 26 '12 at 02:48
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    Can you show how to go from $f(x)=x^{p-1}$ to $f^{-1}(y)=y^{q-1}$ ? - I'm sure it works out somehow using $\frac{1}{p}+\frac{1}{q}=1$, but I'm getting lost in the derivation. – Frank May 26 '12 at 03:23
  • @Frank Make use of the relation $1/p+1/q=1$. This gives us $p=q/(q-1)$ i.e $p-1 = 1/(q-1)$ i.e $1/(p-1)=q-1$ –  May 26 '12 at 03:24
  • @Frank As far as /i recall, I have used this inequality only once to prove the triangle inequality in $L^p$ space. –  May 26 '12 at 03:27
  • @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). –  Sep 18 '12 at 17:02
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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.

Michael Rozenberg
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