1.2.4 Some Algebraic inequalities
Since our study based on some known algebraic inequalities, we
want to recall few of them here. Lemma 1.2.10 ([48],The Cauchy-Schwarz
inequality)
Every inner product satisfies the Cauchy-Schwarz inequality
(x1, X2) kxiM X2M (1.30)
The equality sign holds if and only if X1 and X2 are
dependent.
Young's inequalities :
Lemma 1.2.11 For all a, b 2 R+, we have
ab < 8a2 + b2
48, (1.31)
where 8 is any positive constant.
Proof. Taking the well-known result
(28a -- b)2 ~ 0 Va, b 2 R
for all 8 > 0, we have
482a2 + b2 - 48ab ~ 0.
This implies
48ab < 482a2 + b2
consequently,
ab < 8a2 + 48 1 b2.
This completes the proof.
|
|
Lemma 1.2.12 ([43])
For all a, b ~ 0, the following inequality holds
bq
+ ;
q
a°
ab < p
1 q
= 1.
+
1
where, p
Proof. Let G = (0,1), and f : G --> IR is integrable, such
that
p log a, 0 < x < 1
q log b, 1
P
P
< x < 1
)
8
<>
>:
f(x) =
1 q
= 1.
+
for all a, b > 0 and 1
P
Since cp(t) = et is convex, and using Jensen's
inequality
(,0(1
(1.32)
,u (G) If (x)dx) <
-- ,u (G)
f cp (f (x)) dx.
G G
Consequently, we have
Z
1
~ (G)
G
1
I
0 of (x) dx =
1 /p 1
I
0 ep log adx + I
1/p
cp (f(x)) dx =
eq log bdx
1 / p
I
0
1
I
1/p
=
ap dx +
bq dx
(ap) + (1 -- ) bq .
=
P
(1.33)
G
where, ,u (G) = 1 and
40 ( 1 I
f (x)dx = e(f0 f (x)dx) = e (V/p p log
adx#177;fi1/p q log bdx)
= e(log a#177;log b) = clog ab
= ab. (1.34)
Using (1.32), (1.33) and (1.34) to conclude the result.
|