已知$a,b>0$且$\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{2}{3}$,求$\dfrac{1}{a-1}+\dfrac{4}{b-1}$的最小值.
解:令$m=\dfrac{1}{a},n=\dfrac{1}{b}$,则$m+n=\dfrac{2}{3}$
$\dfrac{1}{a-1}+\dfrac{4}{b-1}=\dfrac{m}{1-m}+\dfrac{4n}{1-n}=\dfrac{1}{1-m}+\dfrac{4}{1-n}-5\ge\dfrac{(1+2)^2}{2-m-n}-5=\dfrac{7}{4}$
练习:
已知$a,b>0$且$\dfrac{1}{a}+\dfrac{1}{b}=2$,求$\dfrac{1}{a+1}+\dfrac{4}{b+1}$的最大值.
答案:$\dfrac{11}{4}$
原文：https://www.cnblogs.com/mathstudy/p/10573777.html