Page 96 - MATINF Nr. 7

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96 PROBLEME DE MATEMATICA PENTRU CONCURSURI

M 123. Ar˘atat ,i c˘a pentru orice n, x, y, z > 0 are loc inegalitatea

n(n + 3)(x + y) + 3 n(n + 3)(y + z) + 3 n(n + 3)(z + x) + 3 27
+ + ≥ 6n + .
x + y + (n + 1)z y + z + (n + 1)x z + x + (n + 1)y (n + 3)(x + y + z)

Daniel Jinga, Pites , ti

Solut ,ia 1. Not˘am x + y + (n + 1)z = a, y + z + (n + 1)x = b s , i z + x + (n + 1)y = c, deci
a + b + c
a, b, c > 0. Avem a + b + c = (n + 3)(x + y + z) s , i a = x + y + z + nz, deci nz = a −
n + 3
(n + 2)a − b − c (n + 2)b − c − a (n + 2)c − a − b
s , i astfel z = . Analog, x = s , i y = . Rezult˘a
n(n + 3) n(n + 3) n(n + 3)
c˘a n(n + 3)(x + y) + 3 = (n + 1)(b + c) − 2a + 3 s , i analoagele, deci inegalitatea din enunt , devine
(n + 1)(b + c) − 2a + 3 (n + 1)(c + a) − 2b + 3 (n + 1)(a + b) − 2c + 3 27
+ + ≥ 6n + ,
a  b ‹  ‹ c a + b + c
b + c c + a a + b 1 1 1 27
adic˘a (n + 1) + + + 3 + + ≥ 6(n + 1) + , care este
a b c a b c a + b + c
b + c c + a a + b b a c
adev˘arat˘a deoarece, folosind Inegalitatea mediilor, avem + + = + + +
a b c a b b
b a c 1 1 1 9
+ + ≥ 2 + 2 + 2 = 6 s , i + + ≥ .
c c a a b c a + b + c
Solut ,ia 2. (Titu Zvonaru, Com˘anes , ti). Folosind Inegalitatea HM-AM avem

3 3 3 27
+ + ≥ . (1)
x + y + (n + 1)z y + z + (n + 1)x z + x + (n + 1)y (n + 3)(x + y + z)

2
2
2
Aplicˆand Inegalitatea lui Bergstrom s , i inegalitatea x + y + z ≥ xy + yz + zx avem
x + y (x + y) 2
X X
(n + 3) = (n + 3)
2
x + y + (n + 1)z (x + y) + (n + 1)(xz + yz)
4(x + y + z) 2
≥ (n + 3) ·
2
2
2
2(x + y + z ) + 2(xy + yz + zx) + 2(n + 1)(xy + yz + zx)
P 2 P P 2 P
2(n + 3) x + 4(n + 3) xy 2n ( x − xy)
= P P = 6 + P P ≥ 6. (2)
2
2
x + (n + 2) xy x + (n + 2) xy
Prin adunarea inegalit˘at , ii (2) ˆınmult , it˘a cu n cu inegalitatea (1) rezult˘a inegalitatea din enunt , .
M 124. Fie n ∈ N, n ≥ 2 s , i a 1 , a 2 , . . . , a n ∈ R astfel ˆıncˆat 0 < a 1 ≤ a 2 ≤ . . . ≤ a n . Ar˘atat ,i c˘a
 ‹
2
n(n − 1) X 3a − 2a i a j + 3a 2 j 3n(n − 1) a 1 a n 2
i
≤ ≤ + − .
2 (a i + a j ) 2 8 a n a 1 3
1≤i<j≤n
Sorin Ulmeanu, Pites , ti s , i Nicolae Papacu, Slobozia

Solut ,ie (Titu Zvonaru, Com˘anes , ti). Prima inegalitate este adev˘arat˘a pentru orice numere reale
n (n − 1)
a 1 , a 2 , . . . , a n neopuse dou˘a cˆate dou˘a, rezultˆand prin adunarea celor relat , ii de tipul
2
2
2
3a − 2a i a j + 3a 2 j = 2 (a i − a j ) + (a i + a j ) 2 ≥ 1.
i
(a i + a j ) 2 (a i + a j ) 2