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2015 Belarus Team Selection Test
4
(a+b+c)^5 >= 81 (a^2+b^2+c^2)abc for a,b,c>0
(a+b+c)^5 >= 81 (a^2+b^2+c^2)abc for a,b,c>0
Source: 2015 Belarus TST 1.4
November 5, 2020
algebra
inequalities
Problem Statement
Prove that
(
a
+
b
+
c
)
5
≥
81
(
a
2
+
b
2
+
c
2
)
a
b
c
(a+b+c)^5 \ge 81 (a^2+b^2+c^2)abc
(
a
+
b
+
c
)
5
≥
81
(
a
2
+
b
2
+
c
2
)
ab
c
for any positive real numbers
a
,
b
,
c
a,b,c
a
,
b
,
c
I.Gorodnin
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