Question 430318
We know that {{{x^2 + y^2}}} must be the longest side of the triangle (it cannot be {{{2xy}}}, I will demonstrate why afterwards). By the Pythagorean theorem,


{{{(x^2 - y^2)^2 + (2xy)^2 = x^4 - 2x^2y^2 + y^4 + 4x^2y^2 = x^4 + 2x^2y^2 + y^4 = (x^2 + y^2)^2}}}. Therefore this is a right triangle.


The only "assumption" we had to make was that {{{x^2 + y^2 > 2xy}}}, but this is easy to prove. By the AM-GM inequality, {{{(x^2 + y^2)/2 >= sqrt(x^2y^2) = xy}}}. Multiplying both sides by 2, {{{x^2 + y^2 >= 2xy}}}. Equality occurs only when {{{x = y}}}, but this cannot be true, so {{{x^2 + y^2}}} is strictly greater than {{{2xy}}}.