Kantorovich 不等式
[T240301] 证明 Kantorovich 不等式: 设 \(f\in R[0,1]\), 且 \(0<m\le f(x)\le M\), 则有
\[\int_0^1f(x)\mathrm {~d}x\int_0^1\frac{1}{f(x)}\mathrm{~d}x\le\frac{(m+M)^2}{4mM}.
\]
证 注意到
\[\frac{\left[f(x)-m\right]\left[f(x)-M\right]}{f(x)}\le0\Longrightarrow f(x)+\frac{mM}{f(x)}\le m+M
\]
两边积分
\[\int_0^1f(x)\mathrm{~d}x+\int_0^1\frac{mM}{f(x)}\mathrm{~d}x\le m+M
\]
又由均值不等式可知
\[\begin{aligned}
\int_0^1f(x)\mathrm {~d}x\int_0^1\frac{1}{f(x)}\mathrm{~d}x&=\frac{1}{mM}\int_0^1f(x)\mathrm {~d}x\int_0^1\frac{mM}{f(x)}\mathrm{~d}x\\
&\le\frac{1}{mM}\left(\frac{\int_0^1f(x)\mathrm{~d}x+\int_0^1\frac{mM}{f(x)}\mathrm{~d}x}{2}\right)^2\\
&\le\frac{(m+M)^2}{4mM}. \quad\quad \#
\end{aligned}
\]
