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A GENERALIZATION OF THE INEQUALITY OF HÖLDER

Florentin Smarandache

University of New Mexico 200 College Road

Gallup, NM 87301, USA E-mail: smarand@unm.edu

One generalizes the inequality of Hödler thanks to a reasoning by recurrence. As particular cases, one obtains a generalization of the inequality of Cauchy-Buniakovski-Scwartz, and some interesting applications. Theorem: If ai

(k ) ∈R+ and pk ∈]1,+∞[ , i ∈{1,2,...,n} , k ∈{1,2,...,m} , such

that:, 1 2

1 1 1... 1mp p p

+ + + = , then:

ai(k )

k=1

m

∏i=1

n

∑ ≤k=1

m

∏ ai(k )( )pk

i=1

n

∑⎛⎝⎜⎞⎠⎟

1

pk

with m ≥ 2 .

Proof: For m = 2 one obtains exactly the inequality of Hödler, which is true. One

supposes that the inequality is true for the values which are strictly smaller than a certain m .

Then:,

( ) ( ) ( )11

2 2( ) ( 1) ( ) ( ) ( 1) ( )

1 1 1 11 1 1

pk k

p pm m mn n n n pk k m m k m mi i i i i i i

i i i ik k k

a a a a a a a− −

− −

= = = == = =

⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟= ⋅ ⋅ ≤ ⋅ ⋅⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠⎜ ⎟

⎝ ⎠

∑ ∑ ∑ ∑∏ ∏ ∏

where 1

p1

+1

p2

+ ...+1

pm−2

+1

p= 1 and ph > 1 , 1≤ h ≤ m − 2 , p > 1 ;

but

( ) ( ) ( )( ) ( )( )11

1 21 2

( 1) ( ) ( 1) ( )

1 1 1

t tt tn n np p p pm m m m

i i i ii i i

a a a a− −

= = =

⎛ ⎞ ⎛ ⎞⋅ ≤ ⋅⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠∑ ∑ ∑

where 1

t1

+1

t2

= 1 and t1 > 1, t2 > 2 .

It results from it:

2

( ) ( ) ( ) ( )1 1

1 2

1 2( 1) ( ) ( 1) ( )

1 1 1

pt ptn n np p pt ptm m m m

i i i ii i i

a a a a− −

= = =

⎛ ⎞ ⎛ ⎞⋅ ≤ ⋅⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠∑ ∑ ∑

with 1

pt1

+1

pt2

=1

p.

Let us note pt1 = pm−1 and pt2 = pm . Then 1 2

1 1 1... 1mp p p

+ + + = is true and one

has pj > 1 for 1 ≤ j ≤ m and it results the inequality from the theorem. Note: If one poses pj = m for 1 ≤ j ≤ m and if one raises to the power m this

inequality, one obtains a generalization of the inequality of Cauchy-Buniakovski-Scwartz:

ai(k )

k=1

m

∏i=1

n

∑⎛⎝⎜⎞⎠⎟

m

≤ ai(k )( )m

i=1

n

∑k=1

m

∏ .

Application: Let a1,a2 ,b1,b2 ,c1,c2 be positive real numbers. Show that:

(a1b1c1 + a2b2c2 )6 ≤ 8(a16 + a2

6 )(b16 + b2

6 )(c16 + c2

6 ) Solution:

We will use the previous theorem. Let us choose p1 = 2 , p2 = 3 , p3 = 6 ; we will obtain the following:

a1b1c1 + a2b2c2 ≤ (a12 + a2

2 )1

2 (b13 + b2

3)1

3 (c16 + c2

6 )1

6 , or more:

(a1b1c1 + a2b2c2 )6 ≤ (a12 + a2

2 )3(b13 + b2

3)2 (c16 + c2

6 ) , and knowing that

3 3 2 6 61 2 1 2( ) 2( )b b b b+ ≤ +

and that (a1

2 + a22 )3 = a1

6 + a26 + 3(a1

4a22 + a1

2a24 ) ≤ 4(a1

6 + a26 )

since a1

4a22 + a1

2a24 ≤ a1

6 + a26 (because: − a2

2 − a12( )2 a1

2 + a22( )≤ 0 )

it results the exercise which was proposed.