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Greece JBMO TST
2015 Greece JBMO TST
1
(3x+y)(3y+z)(3z+x) \ge 64xyz if x,y,z>0
(3x+y)(3y+z)(3z+x) \ge 64xyz if x,y,z>0
Source: Greece JBMO TST 2015 p1
April 29, 2019
algebra
inequalities
three variable inequality
Problem Statement
If
x
,
y
,
z
>
0
x,y,z>0
x
,
y
,
z
>
0
, prove that
(
3
x
+
y
)
(
3
y
+
z
)
(
3
z
+
x
)
≥
64
x
y
z
(3x+y)(3y+z)(3z+x) \ge 64xyz
(
3
x
+
y
)
(
3
y
+
z
)
(
3
z
+
x
)
≥
64
x
yz
. When we have equality;
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