# another proof of Young inequality

Let

 $F(x)=\int_{0}^{x}\phi(t)dt,\text{ and }G(x)=\int_{0}^{x}\phi^{-1}(t)dt.$

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

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

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

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

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

Title another proof of Young inequality AnotherProofOfYoungInequality 2013-03-22 15:45:38 2013-03-22 15:45:38 a4karo (12322) a4karo (12322) 10 a4karo (12322) Proof msc 26D15