Question

设实数m、n满足m+n=4,求根号m^2+n^2-2m+2n+2的最小值大神们帮帮忙

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.