Page 31 - MATINF Nr. 3

Inegalit˘at , i integrale 31

1 1 1 1
Z Z Z Z
2
2
2
2
≤ 9 · (x + a) d x · f (x) d x − (2 + 3a) f (x) d x = (1 + 3a) 2 f (x) d x.
C−B−S
0 0 0 0
1 R 1
R
1
2
2
Astfel, pentru a = − obt , inem ∆ t ≤ 0, deci (2 + 3a) t −3t (x + a) f (x) d x+ 1 f (x) d x ≥
3 4
0 0
Ç å 2
1 1 1 1
2
R
0, (∀) t ∈ R ⇒ 1 R f (x) d x + 2 R f (x) d x ≥ 3 f (x) d x · R xf (x) d x.
a=− 1 4 0 0 0 0
3
1
R
Aplicat , ia 6. Consider˘am f, g : [0, 1] → R dou˘a funct , ii integrabile, astfel ˆıncˆat g (x) d x = 0
0
1
2
R
s , i g (x) d x 6= 0. Ar˘atat , i c˘a are loc inegalitatea:
0
Ç å 2
1
Ñ é 2 R
1 1 f (x) · g (x) d x
Z Z
2
f (x) d x − f (x) d x ≥ 0 .
1
R 2
0 0 g (x) d x
0
Solut ,ie. Folosind condit , ia din enunt , si inegalitatea C-B-S, putem scrie:
1 1
Z Z
λ · f (x) · g (x) d x = g (x) · (1 + λf (x)) d x, (∀) λ ∈ R, de unde
0 0
2 2
1 1 1 1
Ñ é Ñ é
Z Z Z Z
2
2
λ · f (x) · g (x) d x = g (x) · (1 + λf (x)) d x ≤ g (x) d x · (1 + λf (x)) d x
0 0 0 0
Ç å 2
1
R
f (x) · g (x) d x 1 1
Z Z
2
2
2
⇒ λ · 0 ≤ 1 + 2λ · f (x) d x + λ · f (x) d x
1
R
2
g (x) d x 0 0
0
à å í
Ç 2
1
R
1 f (x) · g (x) d x 1
Z Z
2
⇒ λ 2 f (x) d x − 0 + 2λ · f (x) d x + 1 ≥ 0, (∀) λ ∈ R.
1
R 2
0 g (x) d x 0
0
Ç å 2
1 1 R f(x)·g(x) d x
R
2
Cum f (x) d x − 0 ≥ 0, atunci trinomul de gradul al doilea
2
0 1 R g (x) d x
0
2
à
Ç å í
1
R
1 f (x) · g (x) d x 1
Z Z
2
F (λ) = λ 2 f (x) d x − 0 + 2λ · f (x) d x + 1
1
R
2
0 g (x) d x 0
0