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2005 Slovenia Team Selection Test
6
6
Part of
2005 Slovenia Team Selection Test
Problems
(1)
3\sqrt[3]{\frac{1}{abc} +6(a+b+c) }\le \frac{\sqrt[3]3}{abc}
Source: Slovenia TST 2005 p6
2/15/2020
Let
a
,
b
,
c
>
0
a,b,c > 0
a
,
b
,
c
>
0
and
a
b
+
b
c
+
c
a
=
1
ab+bc+ca = 1
ab
+
b
c
+
c
a
=
1
. Prove the inequality
3
1
a
b
c
+
6
(
a
+
b
+
c
)
3
≤
3
3
a
b
c
3\sqrt[3]{\frac{1}{abc} +6(a+b+c) }\le \frac{\sqrt[3]3}{abc}
3
3
ab
c
1
+
6
(
a
+
b
+
c
)
≤
ab
c
3
3
inequalities
algebra