题目
已知 $a,b > 0$,且$\dfrac 1a+\dfrac 1b=2$,求$\dfrac 1{a+1}+\dfrac 4{b+1}$的最大值.
解析
设 $\dfrac{1}{a}=x , \dfrac{1}{b}=y ,$ 则 $x>0, y>0, x+y=2$ 。
$$
\begin{aligned}
\frac{1}{a+1}+\frac{4}{b+1}&=\frac{1}{\frac{1}{x}+1}+\frac{4}{\frac{1}{y}+1} \\
& =\frac{x}{x+1}+\frac{4 y}{y+1} \\
& =\left(1-\frac{1}{x+1}\right)+\left(4-\frac{4}{y+1}\right) \\
& =5-\left(\frac{1^2}{x+1}+\frac{2^2}{y+1}\right) \\
& \leqslant 5-\frac{(1+2)^2}{x+y+2} \\
& =5-\frac{9}{4} \\
& =\frac{11}{4} \\
\end{aligned}
$$
当且仅当 $(a, b)=(3, \dfrac{3}{5})$时,取等号.
故$\dfrac{1}{a+1}+\dfrac{4}{b+1}$的最大值为$\dfrac{11}{4}$.
注
这里用到了柯西不等式的一种重要变形:$$\dfrac{b_1^2}{a_1}+\dfrac{b_2^2}{a_2}+\cdots+\dfrac{b_n^2}{a_n} \geqslant \dfrac{\left(b_1+b_2+\cdots+b_n\right)^2}{a_1+a_2+\cdots+a_n},$$其中$a_i,b_i>0$($i=1,2,\cdots,n$),等号取得的条件是$$\dfrac{b_1}{a_1}=\dfrac{b_2}{a_2}=\cdots=\dfrac{b_n}{a_n}.$$
练习
已知$a,b > 0$,且$\dfrac 1a+\dfrac 1b=\dfrac 23$,求$\dfrac 1{a-1}+\dfrac 4{b-1}$的最小值.
练习参考答案
$\dfrac 74$