對於任意的 $n$ 維向量 $a = \left \{ x_{1},x_{2},...,x_{n} \right \}$，$b = \left \{ y_{1},y_{2},...,y_{n} \right \}$，$p \geq 1$，有
$$\left ( \sum_{i=1}^{n}|x_{i} + y_{i}|^{p} \right )^{\frac{1}{p}} \leq \left ( \sum_{i=1}^{n}|x_{i}|^{p} \right )^{\frac{1}{p}} + \left ( \sum_{i=1}^{n}|y_{i}|^{p} \right )^{\frac{1}{p}}$$
證明：
$$\sum_{i=1}^{n}|x_{i} + y_{i}|^{p} = \sum_{i=1}^{n}|x_{i} + y_{i}|\cdot |x_{i} + y_{i}|^{p - 1} \leq \sum_{i=1}^{n}(|x_{i}| + |y_{i}|)\cdot |x_{i} + y_{i}|^{p - 1} \\
= \sum_{i=1}^{n}|x_{i}| \cdot |x_{i} + y_{i}|^{p - 1} + \sum_{i=1}^{n}|y_{i}| \cdot |x_{i} + y_{i}|^{p - 1} $$
必然存在一個 $q \geq 1$，使得 $\frac{1}{p} + \frac{1}{q} = 1$，則根據 $Holder$ 不等式有
$$\sum_{i=1}^{n}|x_{i}| \cdot |x_{i} + y_{i}|^{p - 1} + \sum_{i=1}^{n}|y_{i}| \cdot |x_{i} + y_{i}|^{p - 1} \leq \left ( \sum_{i=1}^{n}|x_{i}|^{p} \right )^{\frac{1}{p}} \cdot \left ( \sum_{i=1}^{n}|x_{i} + y_{i}|^{q(p-1)} \right )^{\frac{1}{q}} + \left ( \sum_{i=1}^{n}|y_{i}|^{p} \right )^{\frac{1}{p}} \cdot \left ( \sum_{i=1}^{n}|x_{i} + y_{i}|^{q(p-1)} \right )^{\frac{1}{q}} \\
= \left ( \sum_{i=1}^{n}|x_{i}|^{p} \right )^{\frac{1}{p}} \cdot \left ( \sum_{i=1}^{n}|x_{i} + y_{i}|^{p} \right )^{\frac{1}{q}} + \left ( \sum_{i=1}^{n}|y_{i}|^{p} \right )^{\frac{1}{p}} \cdot \left ( \sum_{i=1}^{n}|x_{i} + y_{i}|^{p} \right )^{\frac{1}{q}} \\
= \left \{ \left ( \sum_{i=1}^{n}|x_{i}|^{p} \right )^{\frac{1}{p}} + \left ( \sum_{i=1}^{n}|y_{i}|^{p} \right )^{\frac{1}{p}} \right \} \cdot \left ( \sum_{i=1}^{n}|x_{i} + y_{i}|^{p} \right )^{\frac{1}{q}}$$
於是有
$$\sum_{i=1}^{n}|x_{i} + y_{i}|^{p}
\leq \left \{ \left ( \sum_{i=1}^{n}|x_{i}|^{p} \right )^{\frac{1}{p}} + \left ( \sum_{i=1}^{n}|y_{i}|^{p} \right )^{\frac{1}{p}} \right \} \cdot \left ( \sum_{i=1}^{n}|x_{i} + y_{i}|^{p} \right )^{\frac{1}{q}}$$
兩邊同除以最后一項，便可得
$$\left ( \sum_{i=1}^{n}|x_{i} + y_{i}|^{p} \right )^{\frac{1}{p}} \leq \left ( \sum_{i=1}^{n}|x_{i}|^{p} \right )^{\frac{1}{p}} + \left ( \sum_{i=1}^{n}|y_{i}|^{p} \right )^{\frac{1}{p}}$$
證畢