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Baltic Way
2000 Baltic Way
16
16
Part of
2000 Baltic Way
Problems
(1)
Square root inequality for the reals a,b,c
Source: Baltic Way 2000
12/17/2010
Prove that for all positive real numbers
a
,
b
,
c
a,b,c
a
,
b
,
c
we have
a
2
−
a
b
+
b
2
+
b
2
−
b
c
+
c
2
≥
a
2
+
a
c
+
c
2
\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}\ge\sqrt{a^2+ac+c^2}
a
2
−
ab
+
b
2
+
b
2
−
b
c
+
c
2
≥
a
2
+
a
c
+
c
2
inequalities
analytic geometry
three variable inequality