Page 98 - RMGO 2

98 Rezolvarea problemelor din num˘arul anterior

Clasa a IX-a

2
2
MGO 21. Rezolvat ,i ˆın R × R ecuat ,ia (x − 3x + 3)(y + 5y + 7) = 3(x − 2)(y + 3).
Stelian Corneliu Andronescu s , i Costel B˘alc˘au, Pites , ti

Solut ,ia 1. Cum x = 2 nu convine, ecuat , ia poate fi rescris˘a sub forma f(x) = g(y),
2
x − 3x + 3 3y + 9
unde f : R \ {2} → R, f(x) = , iar g : R → R, g(y) = .
2
x − 2 y + 5y + 7
Avem Im f = (−∞, −1] ∪ [3, ∞) s , i Im g = [−1, 3]. Rezult˘a c˘a f(x) = g(y) = −1
sau f(x) = g(y) = 3, deci (x, y) ∈ {(1, −4), (3, −2)}.
Solut ,ia 2 (Leonard Giugiuc, Drobeta Turnu Severin). Notˆand x−2 = a s , i y +3 = b,
2
2
ecuat , ia este echivalent˘a cu (a +a+1)(b −b+1) = 3ab. Dac˘a ab ≤ 0, clar nu avem

1 1
solut , ii. Cazul 1: a, b > 0. Obt , inem echivalenta a + + 1 b + − 1 = 3.
a b
1 1
Fie a + = u + 2 s , i b + = v + 2. Clar, u, v ≥ 0 s , i ecuat , ia devine uv + u + 3v = 0,
a b
deci u = v = 0, prin urmare a = b = 1, de unde x = 3, y = −2. Cazul 2: a, b < 0.
Analog ca ˆın Cazul 1 obt , inem a = b = −1, de unde x = 1, y = −4.
MGO 22. Ar˘atat ,i c˘a ˆın orice triunghi ascut ,itunghic ABC are loc inegalitatea
√ √ √ √
2 (a + b + c) cos A cos B cos C ≤ bc cos A + ca cos B + ab cos C

(notat ,iile fiind cele obis , nuite).

Leonard Giugiuc, Drobeta Turnu Severin s , i Cristinel Mortici, Tˆargovis , te
√ √ √
Solut ,ie. Fie bc cos A = x, ca cos B = y, ab cos C = z. Conform Teore-
p √ p
2
2
2
2
mei cosinusului, y + z 2 = a, z + x = b, x + y 2 = c, deci cos A =
x 2 y 2 z 2
.
p , cos B = p , cos C = p
2
2
2
2
2
2
2
2
2
2
2
2
(z + x )(x + y ) (x + y )(y + z ) (y + z )(z + x )
Astfel, inegalitatea din enunt , este echivalent˘a cu
1 1 1 x + y + z
≤ .
p + p + p
2
2
2
2
2
2
2
2
2
2
2
2
(x + y )(y + z ) (y + z )(z + x ) (z + x )(x + y ) 2xyz
1 1
=
Utilizˆand Inegalitatea mediilor avem p 2 2 2 2 ≤ p
(x + y )(y + z ) (2xy)(2yz)
√
xz x + z 1 y + x 1
≤ s , i, analog, p ≤ , p ≤
2
2
2
2
2xyz 4xyz (y + z )(z + x ) 4xyz (z + x )(x + y )
2
2
2
2
z + y
, iar prin adunare obt , inem inegalitatea dorit˘a.
4xyz