若 $p,q > 1$，且 $\frac{1}{p} + \frac{1}{q} = 1$，則對於任意的 $n$ 維向量 $a = \left \{ x_{1},x_{2},...,x_{n} \right \}$，$b = \left \{ y_{1},y_{2},...,y_{n} \right \}$，有
$$\sum_{i = 1}^{n}|x_{i}|\cdot |y_{i}| \leq \left ( \sum_{i=1}^{n}|x_{i}|^{p} \right )^{\frac{1}{p}}\left ( \sum_{i=1}^{n}|y_{i}|^{q} \right )^{\frac{1}{q}}$$
證明：
令 $u = \frac{|x_{i}|}{\left ( \sum_{i=1}^{n}|x_{i}|^{p} \right )^{\frac{1}{p}}}$，$v = \frac{|y_{i}|}{\left ( \sum_{i=1}^{n}|y_{i}|^{q} \right )^{\frac{1}{q}}}$，由楊氏不等式有
$$uv = \frac{|x_{i}|}{\left ( \sum_{i=1}^{n}|x_{i}|^{p} \right )^{\frac{1}{p}}} \cdot \frac{|y_{i}|}{\left ( \sum_{i=1}^{n}|y_{i}|^{q} \right )^{\frac{1}{q}}} \leq \frac{u^{p}}{p} + \frac{v^{q}}{q} = \frac{|x_{i}|^{p}}{p\sum_{i=1}^{n}|x_{i}|^{p} } + \frac{|y_{i}|^{q}}{ q\sum_{i=1}^{n}|y_{i}|^{q}}$$
對於上式兩邊 $i$ 從 $1$ 到 $n$ 做連加得
$$\sum_{i=1}^{n}\frac{|x_{i}|}{\left ( \sum_{i=1}^{n}|x_{i}|^{p} \right )^{\frac{1}{p}}} \cdot \frac{|y_{i}|}{\left ( \sum_{i=1}^{n}|y_{i}|^{q} \right )^{\frac{1}{q}}} \leq \sum_{i=1}^{n}\frac{|x_{i}|^{p}}{p\sum_{i=1}^{n}|x_{i}|^{p} } + \sum_{i=1}^{n}\frac{|y_{i}|^{q}}{ q\sum_{i=1}^{n}|y_{i}|^{q}} = \frac{1}{p} + \frac{1}{q} = 1$$
$$\therefore \sum_{i=1}^{n} uv \leq 1$$
於是有
$$\sum_{i = 1}^{n}|x_{i}|\cdot |y_{i}| \leq \left ( \sum_{i=1}^{n}|x_{i}|^{p} \right )^{\frac{1}{p}}\left ( \sum_{i=1}^{n}|y_{i}|^{q} \right )^{\frac{1}{q}}$$
證畢