Belbachir1 . 47KB Jun 04 2011 12:08:24 AM

A GENERALIZED LAGRANGE IDENTITY AND

CAUCHY-BUNIAKOVSKY INEQUALITY

by

Hacene Belbachir

Abstract.The purpose of this Note is to give an extension to the Lagrange identity [1] and [2], and then an extension to the Cauchy-Bouniakovsky inequality, since the right hand side is positive. Finally, we give an application to polynomials.

1. Principal result

Let a₁,...,a_s;b₁,...,b_s∈; and

(

∑

)(

∑

)

∑

+− +−

= 

    −

= m m k k k m k

k

k

m a b a b

k m

M 1 1

0

2 ( 1)

where

∑

akbm+1−k means

∑

=

= +−

s i i

k m i k ib

a

1 1

Theorem. generalized Lagrange identity

     

= +

−

− = −

=

∑

∑

< <

r m b

a b a b a b a

r m b

a b a

j i

i j j i r i j j i j i

r i j j i

m

2 for )

( ) (

1 2 for )

(

M

2 2

2

Corollary. Generalized Cauchy-Buniakovsky inequality

) 1 2 for

( ;

0

M_2;2r−1 ≥ ∀a_i b_j ∈ℜ m= r−

Remark. For m=2r, we deduce the sign of M_2;m If a_ib_j ≥i& j then M_2;m ≥0:

If a_ib_j ≤i& j then M_2;m ≤0: Pour m=1;2;3;4;...on a:

( )( )

(

)

Lagrange Identity

M_2;1=

∑

a_i2

∑

b_i2 −

∑

a_ib_i 2 → :

( )( ) (

∑

∑

−

∑

)(

∑

)

( )( ) (

∑

∑

+

∑

) (

−

∑

)(

∑

)

= 4 4 2 2 2 3 3

3 ;

2 3 4

M a_i b_i a_ib_i a_ib_i a_ib_i

( )( ) (

∑

∑

+

∑

)(

∑

)

−

(

∑

)(

∑

)

= 5 5 3 2 2 3 4 4

4 ;

2 2 3

M a_i b_i a_ib_i a_ib_i a_ib_i a_ib_i

M

2. Proof of the theorem

We have

( )

∑

∑

∑

∑

∑

∑



     − + + 

     −

= + + ab a b

m m ab

b a m b

am m m m m m m

m 1

1

M_2; 1 1 L

1) if m=2r−1

( )

−

( )(

)(

)

+

+

=

∑

∑

∑

−

∑

∑

=

− −

− −

1

1

2 2

1 2 2

2 2

;

2 1

M

r

k

k r k k k r r

k k r

r

r a b a b a b

( )

( )(

∑

)

∑

−

( )

( )(

∑

)(

∑

)

+ =

− −

−

− _{+ −}

−

+ 2 1

1

2 2

1 2 2

1

2 ₁

1

r

r k

k r k k k r r

k k r

r r

k

r _{a b a b a b}

( )

−

( )(

)(

)

+

+

=

∑

∑

∑

−

∑

∑

=

− −

− 1

1

2 2

1 2 2

2 r ₁

k

k r k k k r r

k k r

r _{b a b a b}

a

( )

( )(

∑

)

∑

−

( )

( )(

∑

)(

∑

)

=

− −

− −

− _{+ −}

−

+ 1

1

2 2

1 2 2 2

1

2 ₁

1

r

l

l r l l l r r

l r l r

r r

k

r _{a b a b a b}

( )

{

( ) ( )

}(

)(

)

_+

  

  

+ −

+

=

∑

∑

∑

−

∑

∑

=

− −

− − −

1

1

2 2

1 2

2 1 2 2

2 r ₁

k

k r k k k r r

k r r

k k r

r _{b a b a b}

a

( )

( )(

_{2 1}

)

2

1 −

∑

−

+ r r r

r

k _{a b}

( )

( )(

)(

)

( )

( )(

₂

)

2

1

1

2 2

2 2

2 ₁

2 1

1

∑

∑

∑

∑

∑

∑

_+ −

  

  

− +

= −

=

−

− r r r

k k r

k

k r k k k r r

k k r

r _{b a b a b a b}

a

( )

( )

( ) ( )

( )

( )

( ) ( )

∑ ∑

_

   

− +     

  

−

= −

=

−

j i

r i j r j i r k r r

k

k i j k r j i r k

k _{a b a b a b a b}

;

2 1

0

2

2 ₁

2 1 1

( )

( ) (

)

( )

( ) (

)

+ 

  

    

  

− +

−

=

∑ ∑

∑

< −

=

−

=

− −

j i

r

k

r

k

k j i k r i j r k k

i j k r j i r

k a b a b a b ab

1

0

1

0

2 2

2 2

( )

( ) (

)

( )

( ) (

)

__ 

 _{− + −}

+ r

i j r j i r k r i j r j i r

k ab a b 2 ab a b

2

2 1

( )( ) (

)

( )( ) (

)

+ 

 

  

 _{− + −}

=

∑ ∑

< − =

−

j i

r

k

r i j r j i r r k i j k r j i r

k ab a b ab a b

1

0

2 2

2

2 1

( )

( ) (

)

    −

+

∑

+ =

− r

r l

r j i l j i r

l ab a b

2

1

1 2 2

(

)

∑

_< − =

j i

r i j j

ib a b

a 2

2) if m=2r

( )

−

( )(

)(

)

+

+

=

∑

∑

∑

∑

∑

=

+ − +

− +

+ r

k

k r k k k r r

k k r

r

r a b a b a b

1

1 2 1

2 2

1 2 1 2 2

;

2 1

M

( )

( )(

∑

)(

∑

)

∑

− + − +

+ =

−

−

+

2 1 2 1

2

1

1 2

1

r k k k r k

r

r k

r k k

b

a

b

a

( )

−

( )(

)(

)

+

+

=

∑

∑

∑

∑

∑

=

+ − +

− +

+ r

k

k r k k k r r

k k r

r

b

a

b

a

b

a

1

1 2 1

2 2

1 2 1 2

1

( )

( )

(

∑

)(

∑

)

∑

−+ −+

+

= −+

+

−

+ 2 1 2 1

1

2 1 2 1

1 r l l l r l

r

r k

r l r l

b a b a

( )

{

( )

( )

}

(

∑

)(

∑

)

∑

∑

∑

− + − +

= −+

+

+ _{+ − −}

= 2 1 2 1

1

2 1 2 2 1

2 1

2 r ₁ r k k k r k k

r l r r k k r

r _{b a b a b}

a

We remark that

( )

2

( )

₂2 ₁

( )

2 1

1 2

1 ) (

2 ₊

+

− = −₊ +

− r

k r

l r r k

r k r

; witch gives:

( )

( )(

∑

)(

∑

)

∑

− + − +

=

+

+ + − −

= 2 1 2 1

0

1 2 2

; 2

1 2

1 ) ( 2 1

M r k k k r k

r

k

r k k

r a b a b

r k r

( )

( )

( ) ( )

∑ ∑

_

   

+ + − −

=

=

+ − +

j i

r

k

k i j k r j i r

k k

b a b

a r

k r

; 0

1 2 1

2 1 2

The transposition of i& j; gives:

( )

( ) (

)

∑ ∑

_{< =} + − +

    + − + + − = j i r k k i j k r j i r k

r ab a b

r k r 0 1 2 1 2 2 ; 2 1 2 1 ) ( 2 M

( )

( ) (

)

    − + + − +

∑

= + − + r k k j i k r i j r

k a b ab

r k r 0 1 2 1 2 1 2 1 ) ( 2

( )( ) (

)

∑ ∑

_{< =} + − +







−

+

+

+

−

=

j i r k k i j k r j i r k

r

a

b

a

b

r

k

r

0 1 2 1 2 2 ; 2

1

2

1

)

(

2

M

( )( ) (

−

)



















+

+

−

−

+

∑

+ = + − + r r l l r j i l i j r

l

a

b

a

b

r

l

r

2 1 1 2 1 2

1

2

1

)

(

2

( )( ) (

)

∑ ∑

_{< =}+ + − +













−

+

+

−

=

j i r k k i j k r j i r

k

a

b

a

b

r

k

r

1 2 0 1 2 1 2

1

2

1

)

(

2

let

p

=

a

j

b

i &

q

=

−

a

i

b

j, we have:

( )

∑ ∑

_{< =}+ + − +

        + + − = j i r k k k r r k

r p q

r k r 1 2 0 1 2 1 2 2 ; 2 1 2 1 ) ( 2 M

( )

( )

∑ ∑

_{< =}+ + − +

∑

₌+ + − +













+

−

+

+

−

=

j i r k r k k k r r k k k r r

k

k

p

q

r

q

p

r

k

r

1 2 0 1 2 1 1 2 1 2 1 2 1 2

1

2

1

2

1

)

(

2

( )

( )

∑ ∑

_{< =}+

∑

₌+ − +

− + − +













+

+

−

=

j i r k r k k k r r k k k r r

k

k

p

q

r

r

q

p

1 2 0 1 2 1 1 2 2 1 1 2 1 2

1

)

1

2

(

2

(

)

( )

∑

_< +

∑

₌ − +













+

−

=

j i r l l l r r l r

q

p

k

q

p

2 1 1 2 2 1 2

2

(

)

(

)

{

}

∑

_{< +} + _{− +}

= j i r r q p q q

p 2 1 2 2

(

) (

)

{

}

∑

_< − +

= j i j i i j r j i i

jb ab a b a b

3. Application to polynomials.

Let P( ) ( )

1

∏

= −

= n

i x i

x

µ

and

∑

=

= n

i n n

T

1 1

µ ( µ_i is the module of µ_i) : For

p i i

x

=

µ

&

y

_i

=

µ

_i q we have

( ) ( ) 0

) 1

( _{1 1}

0

≥ 

    

− + +− + +−

=

∑

kp m k q kq m k p

m

k

k _{T T}

k m

This inequality is true for all p and q such that kp+(m+1-k)q & kq+(m+1-k)p are integers.

We have the following relations for m = 1; 2; 3; 4; 5;…

0 1

m= → T_2pT_2q −T_p2_+q ≥

0 2

2

m= → T_3pT_3q − T_2p+qT_p+2q ≥

0 2

4 3

m= → T_4pT_4q − T_3p+qT_p+3q + T₂2_p+2q ≥

0 3

2 4

m= → T_5pT_5q + T_3p+2qT_2p+3q − T_4p+qT_p+4q ≥

0 10

6 15

5

m= → T_6pT_6q + T_4p+2qT_2p+4q − T_5p+qT_p+5q − T₃2_p+3q ≥

M

References:

[1] Hardy G.H., Littlewood J.E., Polya G., Inequalities. Cambridge Univ. Press,1952.

[2] Mitrinovitch P.S.,Vasic P.M., Analytic Inequalities. Springer Verlag, 1970.

Author:

Hacene Belbachir - U.S.T.H.B./Faculte de Mathematiques, El Alia, B.P. 32,