Page 77 - RMGO 3

Rezolvarea problemelor din num˘arul anterior 77

r r
A p(p − a) p(p − b)
0
0
(p − a) cos = (p − a) . Analog, avem B N = (p − b) s , i
2 bc ac
r
p(p − c) p(p − a)(p − b)(p − c)
0
0
0
0
0
0
0
0
C M = (p−c) . Rezult˘a c˘a A P ·B N ·C M = ·
ab abc
S 3
p h a h b h c
p(p − a)(p − b)(p − c) = = .
abc 8
MGO 62. Fie x, y, z ∈ [0, 2] astfel ˆıncˆat xy + yz + zx + xyz = 4. Determinat ,i
4 − xy 4 − yz 4 − zx
valorile extreme ale expresiei + + .
4 + xy 4 + yz 4 + zx
Leonard Mihai Giugiuc, Romˆania s , i Michael Rozenberg, Israel
Solut ,ia 1 (Anh Duy, Vietnam). Cu substitut , iile
2 − 2a 2 − 2b 2 − 2c
x = , y = , z =
1 + a 1 + b 1 + c
P 4 − xy P a + b
avem a, b, c ∈ [0, 1], a + b + c = 1 s , i S = = . Demonstr˘am c˘a
4 + xy 1 + ab
9 X a + b
≤ ≤ 2.
5 1 + ab
P a + b P a + b
Avem S = ≤ = 2(a + b + c) = 2. Notˆand q = ab + bc + ca s , i
1 + ab 1
9 r(q − 9r) + 3r(3q + 2) − (4q − 1)
r = abc, obt , inem c˘a S = + . Cum q − 9r =
2
5 5(q + r + r + 1)
(a + b + c)(ab + bc + ca) − 9abc ≥ 0 (conform Inegalit˘at ,ii mediilor), rezult˘a c˘a
9 3r(3q + 2) − (4q − 1)
2
S ≥ + , (1). Avem (a + b + c) ≥ 3(ab + bc + ca), deci
2
5 5(q + r + r + 1)
1 1 9 1 1
q ≤ . Cazul 1. 0 ≤ q ≤ . Din (1) rezult˘a c˘a S ≥ . Cazul 2. < q ≤ .
3 4 5 4 3
2
2
2
2 2
Din (a + b + c ) + 6abc(a + b + c) ≥ (a + b + c) (ab + bc + ca) (Inegalitatea lui
Schur pentru t = 2) rezult˘a 6r ≥ (1 − q)(4q − 1), prin urmare din (1) obt , inem
9 (1 − q)(4q − 1)(3q + 2) − 2(4q − 1) 9 q(4q − 1)(1 − 3q) 9
S ≥ + = + ≥ .
2
2
5 10(q + r + r + 1) 5 10(q + r + r + 1) 5
Solut ,ia 2 (Toang Huc Khein, software developer, Marea Britanie). Condit , iile
xy + yz + zx + xyz = 4 s , i x, y, z > 0 implic˘a existent , a a trei variabile a, b, c > 0
astfel ˆıncˆat
2a 2b 2c
x = , y = , z =
b + c c + a a + b
a b b c c a a b c
(deoarece · + · + · + 2 · · · = 1).
b + c c + a c + a a + b a + b b + c b + c c + a a + b
Restrict , iile x, y, z ≤ 2 implic˘a a + b ≥ c, b + c ≥ a, c + a ≥ b. Astfel a, b s , i c sunt
lungimile laturilor unui triunghi (posibil degenerat). Avem
X 4 − xy X a(a + b + c)
= .
4 + xy a(a + b + c) + 2bc