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Cuba MO
2012 Cuba MO
4
sum xz / (x^2 + xy + y^2 + 6z^2) <= 1/3
sum xz / (x^2 + xy + y^2 + 6z^2) <= 1/3
Source: 2012 Cuba MO 2.4
September 18, 2024
algebra
inequalities
Problem Statement
Let
x
,
y
,
z
x, y, z
x
,
y
,
z
be positive reals. Prove that
x
z
x
2
+
x
y
+
y
2
+
6
z
2
+
z
x
z
2
+
z
y
+
y
2
+
6
x
2
+
x
y
x
2
+
x
z
+
z
2
+
6
y
2
≤
1
3
\frac{xz}{x^2 + xy + y^2 + 6z^2} + \frac{zx}{z^2 + zy + y^2 + 6x^2} + \frac{xy}{x^2 + xz + z^2 + 6y^2} \le \frac13
x
2
+
x
y
+
y
2
+
6
z
2
x
z
+
z
2
+
zy
+
y
2
+
6
x
2
z
x
+
x
2
+
x
z
+
z
2
+
6
y
2
x
y
≤
3
1
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