Page 35 - MATINF Nr. 1

Inegalit˘at , i integrale 35

Astfel, inegalitatea de demonstrat rezult˘a din:

å Ç å 2 Ç å Ç å
Ç
a + b a + b (b − a) a + b a + b
0
− a f − f (c 1 ) − b − f

2 2 8 2 2

2 2
2 (b − a) (b − a)
0
0
0
0
−f (c 2 ) (b − a) = |f (c 1 ) + f (c 2 )| ≤ · max |f (x)| .
8 8 4 x∈[0,1]

4. Fie f : [0, ∞) → [0, ∞) o funct , ie integrabil˘a. Ar˘atat , i c˘a:
1 1 1
Z Z Z
2
3
f (x) dx · f (x) dx ≤ f (x) dx.
0 0 0
Solut ,ie. Din inegalitatea lui Schur de gradul al III-lea, avem:
3
3
2
3
2
f (x) + f (y) + f (z) + 3f (x) f (y) f (z) ≥ f (x) f (y) + f (y) f (x)
2
2
2
2
+f (y) f (z) + f (z) f (y) + f (z) f (x) + f (z) f (y) ,
(∀) x, y, z ∈ [0, ∞) , de unde prin integrare ˆın funct , ie de cele trei variabile, obt , inem
Z 1 Ç Z 1 å 3 Z 1 Z 1
2
3
3 f (x) dx + 3 f (x) dx ≥ 6 f (x) dx · f (x) dx
0 0 0 0
1 1 1
Z Z Z
3
2
⇒ 6 f (x) dx · f (x) dx ≤ 6 f (x) dx,
Jensen 0 0 0
ceea ce ˆıncheie demonstrat , ia.
5. Fie f : [0, ∞) → [0, ∞) o funct , ie integrabil˘a pe orice interval [0, a] , a > 0, astfel incˆat
ó
x
t
f (x) ≤ 1 2 s , i R î Ä ä − f (t) dt ≤ 1, (∀) x ∈ (0, ∞) . S˘a se arate c˘a
f
x 0 2
1
Z
f (x) dx ≤ 1.
0
Solut ,ie. Relat , ia din enunt , poate fi scris˘a sub forma
Z x Å Z x ã
2 1
f (t) dt ≤ 1 + f (t) dt , (∀) x > 0
0 2 0
Z x 1 Ç Z 2x å
⇒ f (t) dt ≤ 1 + f (t) dt ,
0 2 0
iar prin induct , ie obt , inem c˘a
x 1 2 x
Z Ç Z n å
n
f (t) dt ≤ 2 − 1 + f (t) dt , (∀) n ≥ 1.
0 2 n 0
Acum, pentru a > 1, putem scrie:
Z a Z 1 Z a Z 1 Z a 1
f (x) dx = f (x) dx + f (x) dx ≤ f (x) dx + dx
0 0 1 0 1 x 2
Z 1 1 Z 1
= f (x) dx + 1 − < 1 + f (x) dx := A.
0 a 0
ˆ
In continuare,
x 1 2 x
Z Ç Z n å
n
f (t) dt ≤ 2 − 1 + f (t) dt
0 2 n 0
n
A + 2 − 1 Z 1
≤ → 1, (∀) x > 0 ⇒ f (t) dt ≤ 1.
2 n n→∞ 0