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Contests
National and Regional Contests
Moldova Contests
JBMO TST - Moldova
2010 Junior Balkan Team Selection Tests - Moldova
7
7
Part of
2010 Junior Balkan Team Selection Tests - Moldova
Problems
(1)
\sqrt{(a - b) (a + c)} +\sqrt{(a - c) (a + b) } \ge 2\sqrt{a^2-bc} isosceles
Source: 2010 Moldova JBMO TST p7
2/25/2021
In the triangle
A
B
C
ABC
A
BC
with
∣
A
B
∣
=
c
,
∣
B
C
∣
=
a
,
∣
C
A
∣
=
b
| AB | = c, | BC | = a, | CA | = b
∣
A
B
∣
=
c
,
∣
BC
∣
=
a
,
∣
C
A
∣
=
b
the relations hold simultaneously
a
≥
m
a
x
{
b
,
c
,
b
c
}
,
(
a
−
b
)
(
a
+
c
)
+
(
a
−
c
)
(
a
+
b
)
≥
2
a
2
−
b
c
a \ge max \{ b, c, \sqrt{bc}\}, \sqrt{(a - b) (a + c)} + \sqrt{(a - c) (a + b) } \ge 2\sqrt{a^2-bc}
a
≥
ma
x
{
b
,
c
,
b
c
}
,
(
a
−
b
)
(
a
+
c
)
+
(
a
−
c
)
(
a
+
b
)
≥
2
a
2
−
b
c
Prove that the triangle
A
B
C
ABC
A
BC
is isosceles.
Geometric Inequalities
isosceles