Question

已知a、b、c都是整数,且a²+3b²+3c²+13＜2ab+4b+12c,则a+b+c=

Answer 1

a²+3b²+3c²+13-(2ab+4b+12c)

=(a^2-2ab+b^2)+(2b^2-4b+2)+(3c^2-12c+12)-1

=(a-b)^2+2(b-1)^2+3(c-2)^2-1