Minkowski's Inequality

Theorem for Sums
Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n$ be real numbers.

Let $p \in \R$ be a real number.


 * 1) Let $p > 1$. Then:
 * $\displaystyle \left({\sum_{k=1}^n \left|{a_k + b_k}\right|^p}\right)^{\frac 1 p} \le \left({\sum_{k=1}^n \left|{a_k}\right|^p}\right)^{\frac 1 p} + \left({\sum_{k=1}^n \left|{b_k}\right|^p}\right)^{\frac 1 p}$


 * 1) Let $p < 1, p \ne 0$. Then:
 * $\displaystyle \left({\sum_{k=1}^n \left|{a_k + b_k}\right|^p}\right)^{\frac 1 p} \ge \left({\sum_{k=1}^n \left|{a_k}\right|^p}\right)^{\frac 1 p} + \left({\sum_{k=1}^n \left|{b_k}\right|^p}\right)^{\frac 1 p}$

Theorem for Integrals
Let $f, g$ be integrable functions in $X \subseteq \R^n$ with respect to the volume element $dV$.


 * 1) Let $p > 1$. Then:
 * $\displaystyle \left({\int_X \left|{f + g}\right|^p \mathrm d V}\right)^{\frac 1 p} \le \left({\int_X \left|{f}\right|^p \mathrm d V}\right)^{\frac 1 p} + \left({\int_X \left|{g}\right|^p \mathrm d V}\right)^{\frac 1 p}$


 * 1) Let $p < 1, p \ne 0$. Then:
 * $\displaystyle \left({\int_X \left|{f + g}\right|^p \mathrm d V}\right)^{\frac 1 p} \ge \left({\int_X \left|{f}\right|^p \mathrm d V}\right)^{\frac 1 p} + \left({\int_X \left|{g}\right|^p \mathrm d V}\right)^{\frac 1 p}$

Proof for Sums
The proof for $p=2$ is straightforward:

The result follows from Order Preserved on Positive Reals by Squaring.

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