若正数$a,b,c$满足$dfrac{b+c}{a}+dfrac{a+c}{b}=dfrac{a+b}{c}+1$,则$dfrac{a+b}{c}$的最小值为______
答案:$dfrac{1+sqrt{17}}{2}$
解:记$x=dfrac{a}{c}>0,y=dfrac{b}{c}>0$则由题意$dfrac{y}{x}+dfrac{x}{y}+dfrac{1}{x}+dfrac{1}{y}=x+y+1$
从而 $x+y=dfrac{y}{x}+dfrac{x}{y}+dfrac{1}{x}+dfrac{1}{y}-1gedfrac{1}{x}+dfrac{1}{y}+1gedfrac{4}{x+y}+1 $
故$x+ygedfrac{1+sqrt{17}}{2}$当$x=y=dfrac{1+sqrt{17}}{4}$时成立.
类似的换元见