# $(a-b)^2 = a^2-2ab+b^2$ $(b-c)^2 = b^2-2bc +c^2$ $(a-c)^2 = a^2-2ac+c^2$ So $(a-b)^2 + (b-c)^2 + (a-c)^2 = 2(a^2+b^2+c^2 - (ab+bc+ac))$ But a sum of squares is always positive, so....