Minkowski's Inequality

= Theorem =

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

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


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

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


 * 1) Let $$p > 1$$. Then $$\left({\int_X \left|{f + g}\right|^p dV}\right)^{\frac 1 p} \le \left({\int_X \left|{f}\right|^p dV}\right)^{\frac 1 p} + \left({\int_X \left|{g}\right|^p dV}\right)^{\frac 1 p}$$.
 * 2) Let $$p > 1, p \ne 0$$. Then $$\left({\int_X \left|{f + g}\right|^p dV}\right)^{\frac 1 p} \ge \left({\int_X \left|{f}\right|^p dV}\right)^{\frac 1 p} + \left({\int_X \left|{g}\right|^p dV}\right)^{\frac 1 p}$$.

= Proof =

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

$$ $$ $$ $$

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

.