已知$x,y>0,\dfrac{1}{x}+\dfrac{2}{y}=1$,求$\dfrac{1}{x+1}+\dfrac{2}{y+1}$的最大值____
解答：令$a=\dfrac{1}{x},b=\dfrac{2}{y}$则$a,b>0,a+b=1$
$\dfrac{1}{x+1}+\dfrac{2}{y+1}=\dfrac{a}{a+1}+\dfrac{2b}{b+2}=3-(\dfrac{1}{a+1}+\dfrac{4}{b+2})\le 3-\dfrac{9}{a+b+3}=\dfrac{3}{4}$