若 $p,q > 1$，且 $frac{1}{p} + frac{1}{q} = 1$，则对于任意的 $n$ 维向量 $a = left { x_{1},x_{2},...,x_{n} ight }$，$b = left { y_{1},y_{2},...,y_{n} ight }$，有
$$sum_{i = 1}^{n}|x_{i}|cdot |y_{i}| leq left ( sum_{i=1}^{n}|x_{i}|^{p} ight )^{frac{1}{p}}left ( sum_{i=1}^{n}|y_{i}|^{q} ight )^{frac{1}{q}}$$
证明：
令 $u = frac{|x_{i}|}{left ( sum_{i=1}^{n}|x_{i}|^{p} ight )^{frac{1}{p}}}$，$v = frac{|y_{i}|}{left ( sum_{i=1}^{n}|y_{i}|^{q} ight )^{frac{1}{q}}}$，由杨氏不等式有
$$uv = frac{|x_{i}|}{left ( sum_{i=1}^{n}|x_{i}|^{p} ight )^{frac{1}{p}}} cdot frac{|y_{i}|}{left ( sum_{i=1}^{n}|y_{i}|^{q} ight )^{frac{1}{q}}} leq frac{u^{p}}{p} + frac{v^{q}}{q} = frac{|x_{i}|^{p}}{psum_{i=1}^{n}|x_{i}|^{p} } + frac{|y_{i}|^{q}}{ qsum_{i=1}^{n}|y_{i}|^{q}}$$
对于上式两边 $i$ 从 $1$ 到 $n$ 做连加得
$$sum_{i=1}^{n}frac{|x_{i}|}{left ( sum_{i=1}^{n}|x_{i}|^{p} ight )^{frac{1}{p}}} cdot frac{|y_{i}|}{left ( sum_{i=1}^{n}|y_{i}|^{q} ight )^{frac{1}{q}}} leq sum_{i=1}^{n}frac{|x_{i}|^{p}}{psum_{i=1}^{n}|x_{i}|^{p} } + sum_{i=1}^{n}frac{|y_{i}|^{q}}{ qsum_{i=1}^{n}|y_{i}|^{q}} = frac{1}{p} + frac{1}{q} = 1$$
$$ herefore sum_{i=1}^{n} uv leq 1$$
于是有
$$sum_{i = 1}^{n}|x_{i}|cdot |y_{i}| leq left ( sum_{i=1}^{n}|x_{i}|^{p} ight )^{frac{1}{p}}left ( sum_{i=1}^{n}|y_{i}|^{q} ight )^{frac{1}{q}}$$
证毕