show that for any real a,b both not zero a^2+ab+ b^2 > 0

proof:
      a^2+ab+b^2 = (a^3-b^3)/(a-b)

without loss of generality let a <= b ( as the expression is symetric)

if a < b

then a^3 < b^3 and both a^3-b^3 and a-b are -ve and so ratio is positive

if a = b then a^2 + ab + b^2 = 3a^2 and is zero only if a = 0 that is a=b = 0

proved