(a^2 + b^2 + 1) - (a + b + ab)
= a^2 + b^2 - ab - a - b + 1
= (1/2) (2a^2 + 2b^2 - 2ab - 2a - 2b + 2)
= (1/2) [(a^2 - 2a + 1) + (b^2 - 2b + 1) + (a^2 - 2ab + b^2)]
= (1/2) [(a-1)^2 + (b-1)^2 + (a-b)^2] >= 0
所以 a^2 + b^2 + 1 >= a + b + ab
当且仅当 a = b = 1 时相等
“^2”表示平方