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Algebra Difficulty 4.5 AIME Prove it Soviet Union

Problem:
The positive reals xx, yy satisfy x3+y3=x−yx^{3} + y^{3} = x - y. Show that x2+y2<1x^{2} + y^{2} < 1.

Solution

Solution:
Since xx, yy are positive, so is x3+y3x^{3} + y^{3}, and hence x>yx > y. So
(x2+y2)(x−y)=(x3−y3)−xy(x−y)<x3−y3=x−y.(x^{2} + y^{2})(x - y) = (x^{3} - y^{3}) - xy(x - y) < x^{3} - y^{3} = x - y.
Hence x2+y2<1x^{2} + y^{2} < 1.

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