Question

试证明,不论m,n为何值,m^2+n^2-2m+2n+4的值都为正数

Answer 1

m+n=4
即
m=4-n
，

代入
m^2+n^2-2m+2n+2

m^2+n^2-2m+2n+2

=(4-n)^2+n^2-2(4-n)+2n+2

=16-8n+n^2+n^2-8+2n+2n+2

=2n^2-4n+10

=2(n^2-2n+1)+8

=2(n-3)^2+8
，

由于
2(n-3)^2≥0，

所以
2(n-3)^2+8≥8，

于是 2(n-3)^2-8的最小值为8
，即m^2+n^2-2m+2n+2的最小值为8，

所以根号m^2+n^2-2m+2n+2的最小值为根号8=2根号2.

Answer 2

m^2+n^2-2m+2n+4
=m^2-2m+1+n^2+2n+1+2
=(m-1)^2+(n+1)^2+2
因为(m-1)^2>=0,(n+1)^2>=0
所以(m-1)^2+(n+1)^2+2>0
即不论m,n为何值,m^2+n^2-2m+2n+4的值都为正数