proof of Minkowski inequality

For $p=1$ the result follows immediately from the triangle inequality, so we may assume $p>1$.

We have

$${|a_k+b_k|}^p=|a_k+b_k|{|a_k+b_k|}^{p-1}\le (|a_k|+|b_k|){|a_k+b_k|}^{p-1}$$

by the triangle inequality. Therefore we have

$${|a_k+b_k|}^p\le |a_k|{|a_k+b_k|}^{p-1}+|b_k|{|a_k+b_k|}^{p-1}$$

Set $q=\frac{p}{p-1}$. Then $\frac{1}{p}+\frac{1}{q}=1$, so by the Hölder inequality we have

$$\sum _{k=0}^n|a_k|{|a_k+b_k|}^{p-1}\le {\left(\sum _{k=0}^n{|a_k|}^p\right)}^{\frac{1}{p}}{\left(\sum _{k=0}^n{|a_k+b_k|}^{(p-1)q}\right)}^{\frac{1}{q}}$$

$$\sum _{k=0}^n|b_k|{|a_k+b_k|}^{p-1}\le {\left(\sum _{k=0}^n{|b_k|}^p\right)}^{\frac{1}{p}}{\left(\sum _{k=0}^n{|a_k+b_k|}^{(p-1)q}\right)}^{\frac{1}{q}}$$

Adding these two inequalities, dividing by the factor common to the right sides of both, and observing that $(p-1)q=p$ by definition, we have

$${\left(\sum _{k=0}^n{|a_k+b_k|}^p\right)}^{1-\frac{1}{q}}\le \frac{{\sum }_{k=0}^n(|a_k|+|b_k|){|a_k+b_k|}^{p-1}}{{\left({\sum }_{k=0}^n{|a_k+b_k|}^p\right)}^{\frac{1}{q}}}\le {\left(\sum _{k=0}^n{|a_k|}^p\right)}^{\frac{1}{p}}+{\left(\sum _{k=0}^n{|b_k|}^p\right)}^{\frac{1}{p}}$$

Finally, observe that $1-\frac{1}{q}=\frac{1}{p}$, and the result follows as required. The proof for the integral version is analogous.

Title	proof of Minkowski inequality
Canonical name	ProofOfMinkowskiInequality
Date of creation	2013-03-22 12:42:14
Last modified on	2013-03-22 12:42:14
Owner	Andrea Ambrosio (7332)
Last modified by	Andrea Ambrosio (7332)
Numerical id	10
Author	Andrea Ambrosio (7332)
Entry type	Proof
Classification	msc 26D15
Related topic	HolderInequality