Question

how to solve (a+b+c)3 a^3b^3c^3

Answer 1

a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)

From this, a3+b3+c3=3abca3+b3+c3=3abc if either

a+b+c=0a+b+c=0 or a=b=c,a=b=c,

since a2+b2+c2−ab−bc−ca=(a−b)2+(b−c)2+(c−a)22a2+b2+c2−ab−bc−ca=(a−b)2+(b−c)2+(c−a)22

We also see that the right hand side is a sum of squares, so it must always be non-negative, so a3+b3+c3≥3abca3+b3+c3≥3abc if and only if a+b+c≥0