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minimum (x+y) when (x+\sqrt{1+x^2})(y+\sqrt{1+y^2}) = p

Source: Czech and Slovak Match 1995 P4

October 1, 2017
minimum valuealgebraSum

Problem Statement

For each real number p>1p > 1, find the minimum possible value of the sum x+yx+y, where the numbers xx and yy satisfy the equation (x+1+x2)(y+1+y2)=p(x+\sqrt{1+x^2})(y+\sqrt{1+y^2}) = p.
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