对于任意的 $n$ 维向量 $a = left { x_{1},x_{2},…,x_{n} ight }$，$b = left { y_{1},y_{2},…,y_{n} ight }$，$p geq 1$，有
$$left ( sum_{i=1}^{n}|x_{i} + y_{i}|^{p} ight )^{frac{1}{p}} leq left ( sum_{i=1}^{n}|x_{i}|^{p} ight )^{frac{1}{p}} + left ( sum_{i=1}^{n}|y_{i}|^{p} ight )^{frac{1}{p}}$$
证明：
$$sum_{i=1}^{n}|x_{i} + y_{i}|^{p} = sum_{i=1}^{n}|x_{i} + y_{i}|cdot |x_{i} + y_{i}|^{p – 1} leq sum_{i=1}^{n}(|x_{i}| + |y_{i}|)cdot |x_{i} + y_{i}|^{p – 1} \
= sum_{i=1}^{n}|x_{i}| cdot |x_{i} + y_{i}|^{p – 1} + sum_{i=1}^{n}|y_{i}| cdot |x_{i} + y_{i}|^{p – 1} $$
必然存在一个 $q geq 1$，使得 $frac{1}{p} + frac{1}{q} = 1$，则根据 $Holder$ 不等式有
$$sum_{i=1}^{n}|x_{i}| cdot |x_{i} + y_{i}|^{p – 1} + sum_{i=1}^{n}|y_{i}| cdot |x_{i} + y_{i}|^{p – 1} leq left ( sum_{i=1}^{n}|x_{i}|^{p} ight )^{frac{1}{p}} cdot left ( sum_{i=1}^{n}|x_{i} + y_{i}|^{q(p-1)} ight )^{frac{1}{q}} + left ( sum_{i=1}^{n}|y_{i}|^{p} ight )^{frac{1}{p}} cdot left ( sum_{i=1}^{n}|x_{i} + y_{i}|^{q(p-1)} ight )^{frac{1}{q}} \
= left ( sum_{i=1}^{n}|x_{i}|^{p} ight )^{frac{1}{p}} cdot left ( sum_{i=1}^{n}|x_{i} + y_{i}|^{p} ight )^{frac{1}{q}} + left ( sum_{i=1}^{n}|y_{i}|^{p} ight )^{frac{1}{p}} cdot left ( sum_{i=1}^{n}|x_{i} + y_{i}|^{p} ight )^{frac{1}{q}} \
= left { left ( sum_{i=1}^{n}|x_{i}|^{p} ight )^{frac{1}{p}} + left ( sum_{i=1}^{n}|y_{i}|^{p} ight )^{frac{1}{p}} ight } cdot left ( sum_{i=1}^{n}|x_{i} + y_{i}|^{p} ight )^{frac{1}{q}}$$
于是有
$$sum_{i=1}^{n}|x_{i} + y_{i}|^{p}
leq left { left ( sum_{i=1}^{n}|x_{i}|^{p} ight )^{frac{1}{p}} + left ( sum_{i=1}^{n}|y_{i}|^{p} ight )^{frac{1}{p}} ight } cdot left ( sum_{i=1}^{n}|x_{i} + y_{i}|^{p} ight )^{frac{1}{q}}$$
两边同除以最后一项，便可得
$$left ( sum_{i=1}^{n}|x_{i} + y_{i}|^{p} ight )^{frac{1}{p}} leq left ( sum_{i=1}^{n}|x_{i}|^{p} ight )^{frac{1}{p}} + left ( sum_{i=1}^{n}|y_{i}|^{p} ight )^{frac{1}{p}}$$
证毕