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2000 Austria Beginners' Competition
2
(a+b)^3 / a^2b >= 27/4
(a+b)^3 / a^2b >= 27/4
Source: 2000 Austria Beginners' Competition p2
October 4, 2022
inequalities
algebra
Problem Statement
Let
a
,
b
a,b
a
,
b
positive real numbers. Prove that
(
a
+
b
)
3
a
2
b
≥
27
4
.
\frac{(a+b)^3}{a^2b}\ge \frac{27}{4}.
a
2
b
(
a
+
b
)
3
≥
4
27
.
When does equality occur?
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