Question

已知a,b,1这三个数中至少有2个不相等,比较(a^2)+(b^2)+1与ab+a+b的大小

Answer 1

[(a^2)+(b^2)+1]-[ab+a+b]

=[(a-1)^2+(b-1)^2+(a-b)^2]/2

因为：a,b,1这三个数中至少有2个不相等

所以,

[(a^2)+(b^2)+1]-[ab+a+b]

=[(a-1)^2+(b-1)^2+(a-b)^2]/2

＞0

[(a^2)+(b^2)+1]＞[ab+a+b]