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min of (3x^2 + 16xy + 15y^2)/(x^2 + y^2) - Puerto Rico TST 2010.4
min of (3x^2 + 16xy + 15y^2)/(x^2 + y^2) - Puerto Rico TST 2010.4
Source:
September 14, 2021
algebra
inequalities
min
Problem Statement
Find the largest possible value in the real numbers of the term
3
x
2
+
16
x
y
+
15
y
2
x
2
+
y
2
\frac{3x^2 + 16xy + 15y^2}{x^2 + y^2}
x
2
+
y
2
3
x
2
+
16
x
y
+
15
y
2
with
x
2
+
y
2
≠
0
x^2 + y^2 \ne 0
x
2
+
y
2
=
0
.
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