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2014 Thailand TSTST
2
Sqrt(sum of a^2+b^2/a+b) \geq sum of sqrt(2ab/3a+b+2c)
Sqrt(sum of a^2+b^2/a+b) \geq sum of sqrt(2ab/3a+b+2c)
Source: 2013 Thailand October Camp Inequalities Exam p2
March 7, 2022
inequalities
Problem Statement
Let
a
,
b
,
c
a, b, c
a
,
b
,
c
be positive real numbers. Prove that
a
2
+
b
2
a
+
b
+
b
2
+
c
2
b
+
c
+
c
2
+
a
2
c
+
a
≥
2
a
b
3
a
+
b
+
2
c
+
2
b
c
3
b
+
c
+
2
a
+
2
c
a
3
c
+
a
+
2
b
.
\sqrt{\frac{a^2+b^2}{a+b}}+\sqrt{\frac{b^2+c^2}{b+c}}+\sqrt{\frac{c^2+a^2}{c+a}}\geq\sqrt{\frac{2ab}{3a+b+2c}}+\sqrt{\frac{2bc}{3b+c+2a}}+\sqrt{\frac{2ca}{3c+a+2b}}.
a
+
b
a
2
+
b
2
+
b
+
c
b
2
+
c
2
+
c
+
a
c
2
+
a
2
≥
3
a
+
b
+
2
c
2
ab
+
3
b
+
c
+
2
a
2
b
c
+
3
c
+
a
+
2
b
2
c
a
.
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