another proof of Young inequality

Let

$$F(x)={\int }_0^x\varphi (t)𝑑t,\text{ and }G(x)={\int }_0^x{\varphi }^{-1}(t)𝑑t.$$

Since ${\varphi }^{-1}$ is strictly increasing, $G$ is strictly convex, hence lies above its supporting line, i.e. for every $c$ and $x\ne c$

$$G(b)>G(c)+G^{\prime }(c)(b-c)=G(c)+{\varphi }^{-1}(c)(b-c).$$

In particular, for $c=\varphi (a)$ we have

$$F(a)+G(b)>F(a)+G(\varphi (a))+a(b-\varphi (a))=ab,$$

because $F(a)+G(\varphi (a))=a\varphi (a)$.

Title	another proof of Young inequality
Canonical name	AnotherProofOfYoungInequality
Date of creation	2013-03-22 15:45:38
Last modified on	2013-03-22 15:45:38
Owner	a4karo (12322)
Last modified by	a4karo (12322)
Numerical id	10
Author	a4karo (12322)
Entry type	Proof
Classification	msc 26D15