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4
sum a/{1 + 2b^3} >= (a^2 + b^2 + c^2 + d^2}/3 if a,b,c,d>=0, a+b+c+d = 3
sum a/{1 + 2b^3} >= (a^2 + b^2 + c^2 + d^2}/3 if a,b,c,d>=0, a+b+c+d = 3
Source: 2017 Romania JBMO TST 2.4
June 27, 2020
inequalities
algebra
Problem Statement
Let
a
,
b
,
c
,
d
a, b, c, d
a
,
b
,
c
,
d
be non-negative real numbers satisfying
a
+
b
+
c
+
d
=
3
a + b + c + d = 3
a
+
b
+
c
+
d
=
3
. Prove that
a
1
+
2
b
3
+
b
1
+
2
c
3
+
c
1
+
2
d
3
+
d
1
+
2
a
3
≥
a
2
+
b
2
+
c
2
+
d
2
3
\frac{a}{1 + 2b^3} + \frac{b}{1 + 2c^3} +\frac{c}{1 + 2d^3} +\frac{d}{1 + 2a^3} \ge \frac{a^2 + b^2 + c^2 + d^2}{3}
1
+
2
b
3
a
+
1
+
2
c
3
b
+
1
+
2
d
3
c
+
1
+
2
a
3
d
≥
3
a
2
+
b
2
+
c
2
+
d
2
When does the equality hold?
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