Page 83 - MATINF Nr. 6

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PROBLEME DE MATEMATICA PENTRU CONCURSURI 83

M 110. Fie k ≥ −1 un num˘ar real fixat s , i fie S mult ,imea triunghiurilor neobtuzunghice.
Determinat ,i

n
2
€p p p Š o
min k(ctg A + ctg B + ctg C) + ctg A + ctg B + ctg C
atunci cˆand 4ABC parcurge mult ,imea S.

Leonard Mihai Giugiuc, Drobeta Turnu Severin

Solut ,ie. Definim

Š 2
€p p p
f k (A, B, C) = k(ctg A + ctg B + ctg C) + ctg A + ctg B + ctg C .
π π π π π π
√
Avem f k , , = (k + 3) 3 s , i f k , , = 2k + 4. Conform problemei M 85 din
3 3 3 4 4 2
MATINF 3/2019, rezolvat˘a ˆın MATINF 5/2020, avem
π π π π π π

f √ (A, B, C) ≥ f √ , , = f √ , , .
1+2 3 1+2 3 1+2 3
3 3 3 4 4 2
ˆ
In virtutea celor de mai sus, clam˘am urm˘atoarea:
√
√ §
¦ © 2k + 4, dac˘a − 1 ≤ k ≤ 1 + 2 3,
min f k (A, B, C) = min (k + 3) 3, 2k + 4 = √ √
(k + 3) 3, dac˘a k ≥ 1 + 2 3.
√
Cazul 1. −1 ≤ k ≤ 1 + 2 3. Din solut , ia problemei M 85, deducem c˘a este suficient s˘a
ar˘at˘am c˘a

p
2
2
2
2
2 2
k(x + y + z ) + (x + y + z) ≥ (2k + 4) x y + y z + z x , ∀x, y, z ≥ 0,
2 2
2 2
fiind chiar suficient s˘a consider˘am urm˘atoarele dou˘a cazuri:
2
(i) x ∈ [0, 1] s , i y = z = 1. Problema se reduce la a ar˘ata c˘a (k + 1)x + 4x + 2k + 4 ≥
√
2 3
2
2
2
(2k + 4) 2x + 1, echivalent cu x [(k + 1) x + 8(k + 1)x − 4(k + 5k + 2)x + 16(k + 2)] ≥ 0,
2
2
care se rescrie x {(k + 1)(x − 1) [(k + 1)x + 2(k + 5)] + (−k + 2k + 11)(x + 2)} ≥ 0, adev˘arat.
2
2
(ii) z = 0. Problema se reduce la a ar˘ata c˘a (k + 1)(x + y ) ≥ 2(k + 1)xy, evident adev˘arat.
√
Cazul 2. k ≥ 1 + 2 3. Tot din solut , ia problemei M 85, deducem c˘a este suficient s˘a ar˘at˘am
c˘a √
2
2
2
2
2 2
2 2
2 2
k(x + y + z ) + (x + y + z) ≥ (k + 3) 3 · p x y + y z + z x , ∀x, y, z ≥ 0,
fiind chiar suficient s˘a consider˘am din nou cazurile:
2
(i) x ∈ [0, 1] s , i y = z = 1. Problema se reduce la a ar˘ata c˘a (k + 1)x + 4x + 2k + 4 ≥
√
√
2
2
2 2
2
(k + 3) 3 · 2x + 1, echivalent cu (x − 1) [(k + 1) x + 2(k + 1)(k + 5)x + k − 2k − 11] ≥ 0,
adev˘arat.
√
(ii) z = 0. Deoarece 2k + 4 ≥ (k + 3) 3, problema se reduce din nou la a ar˘ata c˘a
2
2
(k + 1)(x + y ) ≥ 2(k + 1)xy, adev˘arat.
Demonstrat , ia este astfel ˆıncheiat˘a.