MT【341】换元变形

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}-1\ge\dfrac{1}{x}+\dfrac{1}{y}+1\ge\dfrac{4}{x+y}+1 $
故$x+y\ge\dfrac{1+\sqrt{17}}{2}$当$x=y=\dfrac{1+\sqrt{17}}{4}$时成立.