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National High School Mathematics League
2003 National High School Mathematics League
5
The Minumum Value
The Minumum Value
Source: 2003 National High School Mathematics League, Exam One, Problem 5
March 16, 2020
Problem Statement
If
x
,
y
∈
(
−
2
,
2
)
,
x
y
=
−
1
x,y\in(-2,2),xy=-1
x
,
y
∈
(
−
2
,
2
)
,
x
y
=
−
1
, then the minumum value of
u
=
4
4
−
x
2
+
9
9
−
y
2
u=\frac{4}{4-x^2}+\frac{9}{9-y^2}
u
=
4
−
x
2
4
+
9
−
y
2
9
is
(A)
8
5
(B)
24
11
(C)
12
7
(D)
12
5
\text{(A)}\frac{8}{5}\qquad\text{(B)}\frac{24}{11}\qquad\text{(C)}\frac{12}{7}\qquad\text{(D)}\frac{12}{5}\qquad
(A)
5
8
(B)
11
24
(C)
7
12
(D)
5
12
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