# 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 |