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Moscow Mathematical Olympiad
1950 Moscow Mathematical Olympiad
176
MMO 176 Moscow MO 1950 Aa+Bb+Cc>=(Ab+Ac+Ba+Bc+Ca+Cb)/2
MMO 176 Moscow MO 1950 Aa+Bb+Cc>=(Ab+Ac+Ba+Bc+Ca+Cb)/2
Source:
August 6, 2019
angles
sidelengths
geometric inequality
inequalities
Problem Statement
Let
a
,
b
,
c
a, b, c
a
,
b
,
c
be the lengths of the sides of a triangle and
A
,
B
,
C
A, B, C
A
,
B
,
C
, the opposite angles. Prove that
A
a
+
B
b
+
C
c
≥
A
b
+
A
c
+
B
a
+
B
c
+
C
a
+
C
b
2
Aa + Bb + Cc \ge \frac{Ab + Ac + Ba + Bc + Ca + Cb}{2}
A
a
+
B
b
+
C
c
≥
2
A
b
+
A
c
+
B
a
+
B
c
+
C
a
+
C
b
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