# proof of Minkowski inequality

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

We have

 $|a_{k}+b_{k}|^{p}=|a_{k}+b_{k}||a_{k}+b_{k}|^{p-1}\leq(|a_{k}|+|b_{k}|)|a_{k}+% b_{k}|^{p-1}$

by the triangle inequality. Therefore we have

 $|a_{k}+b_{k}|^{p}\leq|a_{k}||a_{k}+b_{k}|^{p-1}+|b_{k}||a_{k}+b_{k}|^{p-1}$

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

 $\sum_{k=0}^{n}|a_{k}||a_{k}+b_{k}|^{p-1}\leq\left(\sum_{k=0}^{n}|a_{k}|^{p}% \right)^{\frac{1}{p}}\left(\sum_{k=0}^{n}|a_{k}+b_{k}|^{(p-1)q}\right)^{\frac{% 1}{q}}$
 $\sum_{k=0}^{n}|b_{k}||a_{k}+b_{k}|^{p-1}\leq\left(\sum_{k=0}^{n}|b_{k}|^{p}% \right)^{\frac{1}{p}}\left(\sum_{k=0}^{n}|a_{k}+b_{k}|^{(p-1)q}\right)^{\frac{% 1}{q}}$

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

 $\left(\sum_{k=0}^{n}|a_{k}+b_{k}|^{p}\right)^{1-\frac{1}{q}}\leq\frac{\sum_{k=% 0}^{n}(|a_{k}|+|b_{k}|)|a_{k}+b_{k}|^{p-1}}{\left(\sum_{k=0}^{n}|a_{k}+b_{k}|^% {p}\right)^{\frac{1}{q}}}\leq\left(\sum_{k=0}^{n}|a_{k}|^{p}\right)^{\frac{1}{% p}}+\left(\sum_{k=0}^{n}|b_{k}|^{p}\right)^{\frac{1}{p}}$

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 ProofOfMinkowskiInequality 2013-03-22 12:42:14 2013-03-22 12:42:14 Andrea Ambrosio (7332) Andrea Ambrosio (7332) 10 Andrea Ambrosio (7332) Proof msc 26D15 HolderInequality