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$\dfrac{1}{3a+4b}+\dfrac{1}{a+3b}=\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{5}(\dfrac{1}{x}+\dfrac{1}{y})(x+2y)\ge\dfrac{3+2\sqrt{2}}{5}$
或者待定系数后利用柯西不等式得
$\dfrac{1}{3a+4b}+\dfrac{1}{a+3b}=\dfrac{1}{3a+4b}+\dfrac{2}{2(a+3b)}\ge\dfrac{(1+\sqrt{2})^2}{5a+10b}=\dfrac{3+2\sqrt{2}}{5}$