Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.9 AIME Prove it Ukraine

Positive numbers $x$ and $y$ satisfy equation $x^2 + y^2 + \frac{8xy}{x+y} = 16$ . Prove that $x + y = 4$ .

Solution

$(x^2 + y^2)(x + y) + 8xy = 16(x + y),$
$( (x + y)^2 - 2xy )(x + y) - 16(x + y) + 8xy = 0,$
$(x + y)^3 - 16(x + y) - 2xy(x + y) + 8xy = 0,$
$(x + y)((x + y)^2 - 16) - 2xy(x + y - 4) = 0,$
$(x + y)(x + y - 4)(x + y + 4) - 2xy(x + y - 4) = 0,$
$(x + y - 4)((x + y)(x + y + 4) - 2xy) = 0.$

For $x > 0$ and $y > 0$
$(x + y)(x + y + 4) - 2xy = x^2 + y^2 + 4(x + y) > 0.$

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