If positive $ a, b, c $ satisfy $ \ dfrac {b + c} {a} + \ dfrac {a + c} {b} = \ dfrac {a + b} {c} + 1 $, the $ \ minimum dfrac {a + b} {c} $ is ______
Answer: $ \ dfrac {1+ \ sqrt {17}} {2} $
Solution: note $ x = \ dfrac {a} {c}> 0, y = \ dfrac {b} {c}> 0 $ by the meaning of problems $ \ dfrac {y} {x } + \ dfrac {x} {y} + \ dfrac {1} {x} + \ dfrac {1} {y} = x + y + 1 $
so $ 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 $
so $ x + y \ ge \ dfrac {1+ \ sqrt {17}} {2} $ when $ x = y = \ dfrac {1+ \ sqrt {17}} {4} $ established time.