Cho $f,g$ liên tục trên $[a,b]$ và $g(x_{0})\neq 0,x_{0}\in [a,b]$Chứng minh rằng:Nếu: $\begin{cases} 0

Đặt: $\begin{cases} \varphi=|g|^{-p},\psi=|fg|^{p}\\ \overline{p}=\frac{1}{p}>1,\overline{q}=\frac{\overline{p}}{\overline{p}-1}=\frac{\frac{1}{p}}{\frac{1}{p}-1}=\frac{1}{1-p}=-\frac{q}{p}>1\end{cases}$
Ta chứng minh được:
$\int\limits^{b}_{a}\varphi(x)\psi(x)dx\leq (\int\limits^{b}_{a}|\varphi(x)|^{\overline{q}}dx)^{\frac{1}{\overline{q}}}.(\int\limits^{b}_{a}|\psi(x)|^{\overline{p}}dx)^{\frac{1}{\overline{p}}}$
$\Rightarrow \int\limits^{b}_{a}|f(x)|^{p}dx\leq (\int\limits^{b}_{a}|g(x)|^{q}dx)^{\frac{-p}{q}}(\int\limits^{b}_{a}|f(x)g(x)|dx)^{p}$
$\Rightarrow \int\limits^{b}_{a}|f(x)|^{p}dx (\int\limits^{b}_{a}|g(x)|^{q}dx)^{\frac{p}{q}}\leq (\int\limits^{b}_{a}|f(x)g(x)|dx)^{p}$
$\Rightarrow \int\limits^{b}_{a}|f(x)g(x)|dx \geq (\int\limits^{b}_{a}|f(x)|^{p}dx)^{\frac{1}{p}}.(\int\limits^{b}_{a}|g(x)|^{q}dx)^{\frac{1}{q}}$
$\Rightarrow$ (ĐPCM)