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2006 Turkey Team Selection Test
3
xy+yz+zx=1 implies inequality
xy+yz+zx=1 implies inequality
Source: Turkey, TST D1, P3
May 10, 2006
inequalities
inequalities proposed
algebra
Problem Statement
If
x
,
y
,
z
x,y,z
x
,
y
,
z
are positive real numbers and
x
y
+
y
z
+
z
x
=
1
xy+yz+zx=1
x
y
+
yz
+
z
x
=
1
prove that
27
4
(
x
+
y
)
(
y
+
z
)
(
z
+
x
)
≥
(
x
+
y
+
y
+
z
+
z
+
x
)
2
≥
6
3
.
\frac{27}{4} (x+y)(y+z)(z+x) \geq ( \sqrt{x+y} +\sqrt{ y+z} + \sqrt{z+x} )^2 \geq 6 \sqrt 3.
4
27
(
x
+
y
)
(
y
+
z
)
(
z
+
x
)
≥
(
x
+
y
+
y
+
z
+
z
+
x
)
2
≥
6
3
.
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