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ARTICOLE SI NOTE DE MATEMATICA
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In leg˘atur˘a cu Problema 4093 din Crux
Mathematicorum

Marin Chirciu 1

Aplicat , ia 1 (Problema 4093 din Crux Mathematicorum, Vol. 41, No. 10 (Dec.2015) - propus˘a
de Dragoljub Milosevi´c). Let ABC be an arbitrary triangle. Let r and R be the inradius and
the circumradius of ABC, respectively. Let m a be the length of the median from vertex A to
side BC and let wa be the length of the internal bisector of ^A to side BC. Define m b ; m c ; w b
and w c similarly. Prove that

a 2 b 2 c 2  R ‹
+ + ≤ 4 − 1 .
m a w a m b w b m c w c r
Dragoljub Milosevi´c, Serbia

Solut ,ie. Folosim inegalitatea lui L. Panaitopol m a l a ≥ p (p − a). (Notat , ie w a = l a ).

X a 2 X a 2 4 (R − r) X a 2 4p (R − r)
Obt , inem: ≤ = , care rezult˘a din: = ,
m a l a p (p − a) r p − a r
X 2
2
X a 2 a (p − b) (p − c) 4p r (R − r) 4 (R − r) Q
2
adev˘arat˘a din = Q = = , (p − a) = r p
2
p − a (p − a) r p r
X
2
2
s , i a (p − b) (p − c) = 4p r (R − r) . Egalitatea are loc dac˘a s , i numai dac˘a triunghiul este
echilateral.
S˘a determin˘am o inegalitate de semn contrar.
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Aplicat , ia 2. In 4ABC
a 2 b 2 c 2  4r ‹ 2
+ + ≥ .
m a w a m b w b m c w c R
Marin Chirciu, Pites , ti

X
2
2
2
Solut ,ie. Folosim inegalitatea lui Bergstr¨om m a l a ≤ m , inegalitatea lui Leibniz a ≤ 9R s , i
a
2
2
inegalitatea lui Mitrinovi´c p ≥ 27r . Obt , inem:
X a 2 X a 2 (a + b + c) 2 4p 2 16p 2
≥ ≥ = = X
2
2
2
2
2
m a l a m 2 m + m + m 2 3 (a + b + c ) 3 a 2
a a b c 4
16p 2 16p 2 16 · 27r 2 16r 2
≥ = ≥ = .
3 · 9R 2 27R 2 27R 2 R 2
Egalitatea are loc dac˘a s , i numai dac˘a triunghiul este echilateral.
Se poate scrie dubla inegalitate:
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Profesor, Colegiul Nat , ional ,,Zinca Golescu”, Pites , ti, marin.chirciu@yahoo.com
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