Page 30 - RMGO 5

30 Rezolvarea problemelor din num˘arul anterior

(xy + yz + zx) 2 (xy + yz + zx) 2
2
2
2
deci xy + yz + zx ≥ . Cum din (1) avem >
x + y + z x + y + z
(xy + yz + zx) 2 2 2 2
a
a
= xy + yz + zx, rezult˘ c˘ xy + yz + zx < xy + yz + zx , deci
xy + yz + zx
2
2
2
xy + yz + zx + xyz < xy + yz + zx + xyz ≤ 4.
a
Ar˘at˘am ˆın continuare c˘ dac˘ x, y, z ≥ 0 verific˘ inegalit˘t , ile xy + yz + zx > 3
a
a
a
s , i xy + yz + zx + xyz < 4, atunci x + y + z > xy + yz + zx.
Putem presupune, f˘ar˘ a restrˆange generalitatea, c˘ z ≥ y ≥ x.
a
a
2
Din xy + yz + zx > 3 s , i yz < 4 avem 1 < yz < 4. Not˘am y + z = 2s s , i yz = p .
4 − p 2
Atunci 1 < p < 2 s , i p ≤ s. Avem x (y + z + yz) < 4 − yz, deci x < . Avem
2s + p 2
2
de demonstrat c˘a xy + yz + zx − (x + y + z) ≤ 0, adic˘a x (2s − 1) − 2s + p ≤ 0.
2
4 − p
2
Fix˘am s s , i p s , i consider˘am funct , ia f (t) = t (2s − 1) − 2s + p , t ∈ 0, .
2s + p 2
Deoarece 2s − 1 > 0, avem
4 − p (4 − p )(2s − 1)
2 2
2
max f(t) = f = − 2s + p .
h i 2s + p 2 2s + p 2
t∈ 0, 4−p 2 )
2s+p 2
2 2
4 − p (4 − p )(2s − 1)
2
Astfel xy+yz+zx−(x+y+z) = f(x) < f = −2s+p .
2s + p 2 2s + p 2
4 − p 2 (2t − 1)
2
Fix˘am p s , i consider˘am funct , ia g (t) = − 2t + p , t ∈ [p, ∞).
2t + p 2
2
0
g (t) 4 − p 2 p + 1
Avem = − 1, pentru orice t ∈ [p, ∞).
2 2
2 (2t + p )
2
4 − p 2 p + 1
2
2
2
Cum 0 < 4 − p < 3 s , i 2t + p ≥ 2p + p , avem − 1 <
2 2
(2t + p )
2
2
3 p + 1 3 p + 1
− 1, pentru orice t ∈ [p, ∞) . Ar˘at˘am c˘a − 1 < 0, adic˘a
2 2
2 2
(2p + p ) (2p + p )
ˆ
2
4
0
3
3 < p + 4p + p , adev˘arat. In concluzie avem g (t) < 0 pentru orice t ∈ [p, ∞),

(2 − p)(2p − 1) 4 − p 2 (2s − 1)
2
deci max g(t) = g(p) = − 2p + p . Astfel −
t∈[p,∞) p 2s + p 2
(2 − p) (2p − 1)
2
2
2s + p = g (s) ≤ g (p) = − 2p + p . Ar˘at˘am c˘a g(p) < 0, adic˘a
p
(2 − p) (2p − 1) 2
< p (2 − p), adic˘a 0 < (p − 1) , adev˘arat.
p
Conform inegalit˘at , ilor demonstrate, rezult˘a c˘a xy + yz + zx − (x + y + z) <
4 − p 2 (2s − 1) (2 − p) (2p − 1)
2
2
−2s+p ≤ −2p+p < 0, prin urmare am obt , inut
2s + p 2 p
x + y + z > xy + yz + zx, contradict , ie cu (1). Demonstrat , ia este complet˘a.