Question

设实数x，y满足x^2+(y-1)^2=1，若对满足x,y,不等式x+y+c≥0恒成立 ，则c的取值范围

Answer 1

x=cosa y=1+sina
x+y=cosa+1+sina=1+根2sin(a+派/4)
x+y的最小值为1-根2
x+y≥-c
1-根2≥-c
c≥根2-1

Answer 2

换句话即x+y的取值范围设x+y=t x=t-y
t*t-2ty+y*y+y*y-2y+1=1
即2y*y-（2t+2）y+t*t=0
判别式》=0解出1-根号2《=t《=1+根号2
x+y+c≥0 c》=-（x+y） c>=（根号2）-1

Answer 3

由柯西不等式，(x+y-1)^2<=2[(x^2+(y-1)^2]=2
x+y-1>=-√2
x+y>=1-√2
因为x+y>=-c恒成立，故-c<=1-√2，即c>=√2-1