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1+2abc>=a^2+b^2+c^2 => 1+2(abc)^2>=a^4+b^4+c^4 (HOMC 2015 J Q6)
1+2abc>=a^2+b^2+c^2 => 1+2(abc)^2>=a^4+b^4+c^4 (HOMC 2015 J Q6)
Source:
August 6, 2019
algebra
inequalities
Problem Statement
Let
a
,
b
,
c
∈
[
−
1
,
1
]
a, b, c \in [-1, 1]
a
,
b
,
c
∈
[
−
1
,
1
]
such that
1
+
2
a
b
c
≥
a
2
+
b
2
+
c
2
1 + 2abc \ge a^2 + b^2 + c^2
1
+
2
ab
c
≥
a
2
+
b
2
+
c
2
. Prove that
1
+
2
a
2
b
2
c
2
≥
a
4
+
b
4
+
c
4
1 + 2a^2b^2c^2 \ge a^4 + b^4 + c^4
1
+
2
a
2
b
2
c
2
≥
a
4
+
b
4
+
c
4
.
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