[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
\[\begin{aligned} \int_0^1f(x)\mathrm {~d}x\int_0^1\frac{1}{f(x)}\mathrm{~d}x&=\frac{1}{mM}\int_0^1f(x)\mathrm {~d}x\int_0^1\frac{mM}{f(x)}\mathrm{~d}x\\ &\le\frac{1}{mM}\left(\frac{\int_0^1f(x)\mathrm{~d}x+\int_0^1\frac{mM}{f(x)}\mathrm{~d}x}{2}\right)^2\\ &\le\frac{(m+M)^2}{4mM}. \quad\quad \# \end{aligned} \] 标签：le,frac,不等式,int,mM,1f,Kantorovich,mathrm From： https://www.cnblogs.com/hznudmh/p/18046597