I was so foolish

0<a<1

0<b<1

0<c<1

c=1-a-b

0<1-a-b<1

-1<-a-b<0

1>a+b>0

since 0<a, 0<b, there must be 0<a+b,

so we only need to test whether a+b<1 is satisified.

原文地址:https://www.cnblogs.com/len3d/p/1332931.html