Iranian Mathematical Olympiad

Because of the problem's assumption, it is enough to prove that $$(\prod (x-y){)}^2(\sum x^2)\le \sum (x^2-y^2{)}^2(\sum (xy{)}^2).$$ By Cauchy-Schwarz inequality we have $$(\sum xy(x^2-y^2){)}^2\le \sum (x^2-y^2{)}^2(\sum (xy{)}^2).(1)$$ On the other hand, an easy calculation shows that $$(\prod (x-y){)}^2(\sum x{)}^2=(\sum xy(x^2-y^2){)}^2.$$ Finally, we have $$(\prod (x-y){)}^2(\sum x^2)\le (\prod (x-y){)}^2(\sum x{)}^2=(\sum xy(x^2-y^2){)}^2.(2)$$