Page 130 - MATINF Nr. 13-14

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130 PROBLEME DE MATEMATICA PENTRU CONCURSURI

Clasa a X-a

M 210. Fie a, b, c, d, e, f numere reale nenegative astfel ˆıncˆat

ab + bc + cd + de + ef + fa = 6.

a
Demonstrat ,i c˘
2
2
2
2
2
2
(2a + 1) + (2b + 1) + (2c + 1) + (2d + 1) + (2e + 1) + (2f + 1) ≥ 54.
Vasile Cˆırtoaje, Ploies , ti
2
Solut ,ie. Fie s = a + c + e s , i q = ac + ce + ea, deci 3q ≤ s .
Aplicˆand Inegalitatea Cauchy-Schwarz, avem

2 2 2 2 2 2
(a + c) + (c + e) + (e + a) (2b + 1) + (2d + 1) + (2f + 1)
2
≥ [(a + c)(2b + 1) + (c + e)(2d + 1) + (e + a)(2f + 1)]
2
2
= 4(a + c + e + ab + bc + cd + de + ef + fa) = 4(s + 6) .
Deoarece

2
2
2
2
2
2
2
(a + c) + (c + e) + (e + a) = 2(a + c + e + ac + ce + ea) = 2(s − q) > 0,
rezult˘
a
2(s + 6) 2
2
2
2
(2b + 1) + (2d + 1) + (2f + 1) ≥ .
s − q
2
a
a
Astfel, este suficient s˘ ar˘at˘am c˘
2(s + 6) 2
2
2
2
(2a + 1) + (2c + 1) + (2e + 1) + ≥ 54.
2
s − q
T , ˆınˆand seama c˘a
2
2
2
2
2
2
2
(2a + 1) + (2c + 1) + (2e + 1) = 4(a + c + e + a + c + e) + 3 = 4(s + s − 2q) + 3,
a
r˘amˆane s˘ demonstr˘am c˘a
2(s + 6) 2
2
4(s + s − 2q) + ≥ 51,
2
s − q
2
2
3
2
4
a
adic˘ f(q) ≥ 0, unde f(q) = 8q − (12s + 4s − 51)q + 4s + 4s − 49s + 24s + 72.
2
2
2
Pentru s ≤ 1, avem f(q) > 8q − (12s + 4s − 51)q ≥ (51 − 4s − 12s )q ≥ 0.
Consider˘am acum s ≥ 1 s , i scriem funct , ia f(q) sub forma
Å 2 ã 2
12s + 4s − 51 g(s)
f(q) = 8 − q + ,
16 32
4
3
2
unde g(s) = −16s + 32s − 360s + 1176s − 297.