Question

设a,b为实数,求a2+2ab+b2-4b+5的最小值，并求此时a与b的值

Answer 1

应该是a²+2ab+2b²-4b+5
=(a²+2ab+b²)+(b²-4b+4)+1
=(a+b)²+(b-2)²+1
所以b=2,a=-b=-2时，最小值是1

Answer 2

a^2+2a+b^2-4b+5
=a^2+2a+1+b^2-4b+4
=(a+1)^2+(b-2)^2
当两式相等时
a+1=0
a=-1
b=2

最小值-1+2=1