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All-Soviet-Union Mathematical Olympiad

Algebra Difficulty 4.5 AIME Prove it Soviet Union

Problem:
The positive reals $x$ , $y$ satisfy $x^{3} + y^{3} = x - y$ . Show that $x^{2} + y^{2} < 1$ .

Solution

Solution:
Since $x$ , $y$ are positive, so is $x^{3} + y^{3}$ , and hence $x > y$ . So
$(x^{2} + y^{2})(x - y) = (x^{3} - y^{3}) - xy(x - y) < x^{3} - y^{3} = x - y.$
Hence $x^{2} + y^{2} < 1$ .

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