Page 94 - REVISTA MATINF Nr. 5

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94 PROBLEME DE MATEMATICA PENTRU CONCURSURI

M 100. Fie f, g : [a, b] → R dou˘a funct ,ii derivabile cu derivatele continue astfel ˆıncˆat
2
2
f (x) + g (x) 6= 0, oricare ar fi x ∈ [a, b]. Demonstrat ,i c˘a
s
Ê
b 0 0 2 2
Z 2 2
(f (x)) + (g (x)) f (b) + g (b)
dx ≥ ln .
2
2
2
2
f (x) + g (x) f (a) + g (a)
a
Cristinel Mortici, Viforˆata
Solut ,ie (Alexandru Daniel Pˆırvuceanu, elev, Drobeta Turnu Severin; Leonard Giugiuc, Drobeta
Turnu Severin). Conform Inegalit˘at ,ii Cauchy-Buniakowski-Schwarz avem
È
0
0
0
0
2
2
2
[(f (x)) + (g (x)) ] · [f (x) + g (x)] ≥ f(x)f (x) + g(x)g (x), ∀x ∈ [a, b].
2
ˆ 2 2
Imp˘art , ind aceast˘a inegalitate cu f (x) + g (x) > 0, ∀x ∈ [a, b], obt , inem
Ê
0
0
0
2
2
0
(f (x)) + (g (x)) 2 f(x)f (x) + g(x)g (x) 1 [f (x) + g (x)] 0
2
≥ = · , ∀x ∈ [a, b].
2
2
2
2
2
2
f (x) + g (x) f (x) + g (x) 2 f (x) + g (x)
Prin integrarea ultimei inegalit˘at , i pe [a, b] obt , inem
Ê Ê
(f (x)) + (g (x)) 1 f (b) + g (b)
Z b 0 2 0 2 b 2 2
2
2
dx ≥ ln(f (x) + g (x)) = ln .
2
2
2
f (x) + g (x) 2 f (a) + g (a)
2
a a