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.


 * $(1):\quad$ Let $p > 1$. Then:
 * $\displaystyle \left({\sum_{k=1}^n \left\vert{a_k + b_k}\right\vert^p}\right)^{1/p} \le \left({\sum_{k=1}^n \left\vert{a_k}\right\vert^p}\right)^{1/p} + \left({\sum_{k=1}^n \left\vert{b_k}\right\vert^p}\right)^{1/p}$
 * $(2):\quad$ Let $p < 1, p \ne 0$. Then:
 * $\displaystyle \left({\sum_{k=1}^n \left\vert{a_k + b_k}\right\vert^p}\right)^{1/p} \ge \left({\sum_{k=1}^n \left\vert{a_k}\right\vert^p}\right)^{1/p} + \left({\sum_{k=1}^n \left\vert{b_k}\right\vert^p}\right)^{1/p}$

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


 * $(1):\quad$ Let $p > 1$. Then:
 * $\displaystyle \left({\int_X \left\vert{f + g}\right\vert^p \mathrm d V}\right)^{1/p} \le \left({\int_X \left\vert{f}\right\vert^p \mathrm d V}\right)^{1/p} + \left({\int_X \left\vert{g}\right\vert^p \mathrm d V}\right)^{1/p}$
 * $(2):\quad$ Let $p < 1, p \ne 0$. Then:
 * $\displaystyle \left({\int_X \left\vert{f + g}\right\vert^p \mathrm d V}\right)^{1/p} \ge \left({\int_X \left\vert{f}\right\vert^p \mathrm d V}\right)^{1/p} + \left({\int_X \left\vert{g}\right\vert^p \mathrm d V}\right)^{1/p}$

Proof for $p > 1$
Let $\mathbf{x}$ and $\mathbf{y}$ be members of the Lebesgue space $\ell^p$.

Without loss of generality, assume that $\mathbf{x}$ and $\mathbf{y}$ are non-zero.

Then:

The result follows by dividing both sides of the above inequality by $\left\Vert { \mathbf{x} + \mathbf{y} } \right\Vert_p^{p-1}$.

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

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