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Saudi Arabia Contests
Saudi Arabia JBMO TST
2017 Saudi Arabia JBMO TST
5
1/(a^2b+16)+1/(b^2c+16)+1/(c^2a+16)>=1/8 if a,b,c>0, a+b+c=6$
1/(a^2b+16)+1/(b^2c+16)+1/(c^2a+16)>=1/8 if a,b,c>0, a+b+c=6$
Source: 2017 Saudi Arabia JBMO Training Tests 5
May 28, 2020
inequalities
algebra
Problem Statement
Let
a
,
b
,
c
>
0
a,b,c>0
a
,
b
,
c
>
0
and
a
+
b
+
c
=
6
a+b+c=6
a
+
b
+
c
=
6
. Prove that
1
a
2
b
+
16
+
1
b
2
c
+
16
+
1
c
2
a
+
16
≥
1
8
.
\frac{1}{a^2b+16}+\frac{1}{b^2c+16}+\frac{1}{c^2a+16} \ge \frac{1}{8}.
a
2
b
+
16
1
+
b
2
c
+
16
1
+
c
2
a
+
16
1
≥
8
1
.
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