inequalityreal-analysisyoung-inequality
Let's take a look at Young's inequality: If $u,v\geqslant 0$ and $p,q$ – positive real numbers such that $\frac{1}{p}+\frac{1}{q}=1$ then $$\dfrac{u^p}{p}+\dfrac{v^q}{q}\geqslant uv.$$
It's easy to check that if $u^p=v^q$ then we get equality.
But how to prove that if $\dfrac{u^p}{p}+\dfrac{v^q}{q}=uv$ then $u^p=v^q$?
Can anyone show to do this?
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.
This proof is from "Mathematical Toolchest" published by the Australian Mathematics Trust (image).
Example. If $p$ and $q$ are positive rationals such that $\frac1p + \frac1q = 1$, then for positive $x$ and $y$ $$\frac{x^p}p + \frac{y^q}q \ge xy.$$
Since $\frac1p + \frac1q = 1$, we can write $p = \frac{m+n}m$, $q = \frac{m+n}n$ where $m$ and $n$ are positive integers. Write $x = a^{1/p}$, $y = b^{1/q}$. Then $$\frac{x^p}p + \frac{y^q}q = \frac a{\frac{m+n}m} + \frac b{\frac{m+n}n} = \frac{ma + nb}{m + n}.$$
However, by the AM–GM inequality, $$\frac{ma + nb}{m + n} \ge (a^m \cdot b^n)^{\frac1{m+n}} = a^{\frac1p} b^{\frac1q} = xy,$$ and thus $$\frac{x^p}p + \frac{y^q}q \ge xy.$$
Best Answer
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 $u^p<v^q$.