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min of (3x^2 + 16xy + 15y^2)/(x^2 + y^2) - Puerto Rico TST 2010.4

Source:

September 14, 2021
algebrainequalitiesmin

Problem Statement

Find the largest possible value in the real numbers of the term 3x2+16xy+15y2x2+y2\frac{3x^2 + 16xy + 15y^2}{x^2 + y^2} with x2+y20x^2 + y^2 \ne 0.