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$m+n=dfrac{2}{3}$
$dfrac{1}{a-1}+dfrac{4}{b-1}=dfrac{m}{1-m}+dfrac{4n}{1-n}=dfrac{1}{1-m}+dfrac{4}{1-n}-5gedfrac{(1+2)^2}{2-m-n}-5=dfrac{7}{4}$
练习1:
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:$dfrac{11}{4}$
练习2:
已知$a,b>0,a+2b=1$,则$dfrac{1}{3a+4b}+dfrac{1}{a+3b}$的最小值为_____
解答:令$3a+4b=x,a+3b=y$则$a=dfrac{3x-4y}{5},b=dfrac{3y-x}{5},x+2y=5$
故$dfrac{1}{3a+4b}+dfrac{1}{a+3b}=dfrac{1}{x}+dfrac{1}{y}=dfrac{1}{5}(dfrac{1}{x}+dfrac{1}{y})(x+2y)gedfrac{3+2sqrt{2}}{5}$
或者待定系数后利用柯西不等式得
$dfrac{1}{3a+4b}+dfrac{1}{a+3b}=dfrac{1}{3a+4b}+dfrac{2}{2(a+3b)}gedfrac{(1+sqrt{2})^2}{5a+10b}=dfrac{3+2sqrt{2}}{5}$
练习3:
$dfrac{1}{(2a+b)b}+dfrac{2}{(2b+a)a}=1$求$ab$的最大值
答案：2-$dfrac{2sqrt{2}}{3}$
提示:条件两边同乘$ab$齐次化后分母双代换.