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 a 1 ,..., a s ; b1 ,..., b s ∈ ; and

where

∑ a k b m+1−k

M 2m =

means

∑ (−1) k  k  (∑ a m+1−k b k ) (∑ a k b m+1−k )

m

m

k =0
i = s k m +1− k
a b
i =1 i i

∑

∑

Theorem. generalized Lagrange identity
 (ai b j − a j bi ) 2 r
 i< j
M 2m = 
2r
 (ai b j − a j bi ) ( ai b j + a j bi )
 i< j

∑

for m = 2r − 1
for m = 2r

Corollary. Generalized Cauchy-Buniakovsky inequality
M 2; 2 r −1 ≥ 0 ∀ai ; b j ∈ ℜ (for m = 2r − 1)
Remark. For m = 2r , we deduce the sign of M 2;m

If ai b j ≥ i & j then M 2;m ≥ 0 :

If ai b j ≤ i & j then M 2;m ≤ 0 :
Pour m = 1;2;3;4;... on a:

(∑ a )(∑ b ) − (∑ a b ) → Lagrange Identity :
= (∑ a )(∑ b ) − (∑ a b )(∑ a b )

M 2;1 =

M 2;2

2

2
i

2
i

i i

3
i

3
i

2
i i

1

2
i i

(∑ a
= (∑ a

M 2;3 =

4
i

M 2;4

5
i

M

)(∑ b ) + 3(∑ a b ) − 4(∑ a b )(∑ a b )
)(∑ b ) + 2(∑ a b )(∑ a b ) − 3(∑ a b )(∑ a b )
4
i

2 2 2
i i

5
i

3 2
i i

3
i i

3
i i

2 3
i i

4
i i

4
i i

2. Proof of the theorem

∑ a m+1 ∑ b m+1 −  1 ∑ a m b∑ ab m + L + (− 1)m  m ∑ ab m ∑ a m b

We have

M 2;m =

m

1) if m = 2r − 1

M 2; 2 − r =

+ (− 1)r
=

b 2r +

∑ (− 1)k ( 2rk−1 )(∑ a 2r −k b k )(∑ a k b 2r −k ) +
r −1

( )(∑ a b ) + ∑ (− 1) ( )(∑ a
2 r −1
k

r r 2

k =1
2 r −1

k = r +1

( )(∑ a b ) + ∑ (− 1) ( )(∑ a
k =1

2 r −1
k

r r 2

∑ a 2r ∑ b 2r

∑

r −1

{( ) + (

l =1

 r −1
+  (− 1)k

 k =1

l 2 r −1
2 r −l

2 r −1
k

( )(∑ a b )

= ∑ a ∑ b +  ∑ (− 1) ( )(∑ a


+ (− 1)k
2r

=

k 2 r −1
k

2r −k k

b

)(∑ a b )
k 2r −k

∑ a 2r ∑ b 2r + ∑ (− 1)k ( 2rk−1 )(∑ a 2r −k b k )(∑ a k b 2r −k ) +
r −1

+ (− 1)r
=

∑ ∑
a 2r

m

2 r −1
r

2r

∑ ∑
i; j

r r 2

r −1

k =1

 r −1
k
 (− 1)
 k =0

k 2r
k

2r −k k

b

2 r −1
2r −k

2 r −l l

)}(∑ a

b

)(∑ a b )
l 2 r −l

2r −k k

b

)(∑ a b ) +
k 2r −k

)(∑ a b ) + 12 (− 1) ( )(∑ a b )
k 2r −k

r r 2

k 2r
k



( )(a b ) (a b )  + 12 (− 1) ( )(a b ) (a b ) 
2r
k

i j

2r −k

k

j i

the permutation of i & j gives:

2





r 2r
k

r

i j

r

j i



=


∑ ∑ ( )(a b ) (− a b ) + ∑ ( )(a b ) (− a b )  +

 r −1

i< j 
k =0

2r
k

2r
k

i j

1
+
2

2 r −k

2r
k

( )(a b ) (− a b ) + ( )(a b ) (− a b )  
i j

r


= ∑  ∑
i < j  k =0
r −1

r

( )(a b ) (− a b )

k

j

j i



2r

k

i j

( )(a b ) (− a b )  +
j i

+

2r −k



r

i j

2r −k

i

j i

r

2r
k

j i

2r
k

k =0

j i


∑ (2l r )(ai b j )l (− ai b j )2r −1 
l = r +1

2r
= ∑ (ai b j − a j bi )
+

r −1

k

1
2

r

2r
r

i

r

j

j i



i< j

2) if m = 2r
M 2;2 r =

∑

∑

a 2 r +1

+

b 2r +1 +

∑ (− 1)k (2kr )(∑ a 2r −k +1b k )(∑ a k b 2r −k +1 ) +
r

∑ (− 1) ( )(∑ a
2r

k

k = r +1

k =1

2 r −1
k

= ∑ a 2 r +1 ∑ b 2 r +1 + ∑ (− 1)
r

=

∑

a 2 r +1

We remark that
M 2;2 r =
=

r

k =0

2r
k

)(∑ a b
k

2 r − k +1

bk

2 r − k +1

)(∑ a b
k

)

2 r − k +1

)+

r

k = r +1

b 2 r +1 +

( )− (
2r
k

( )(∑ a

bk

∑ (− 1)l +1 (2r2−rl +1 )(∑ a 2r −l +1b l )(∑ a l b 2r −l +1 )
k =1

∑ (− 1)k {( 2kr ) − (2r2−rl +1 )}(∑ a 2r −k +1b k )(∑ a k b 2r −k +1 )
r

k =1

2r
2 r −l +1

∑ (− 1)k

∑∑
i; j

∑

+

k

2 r − k +1

) = 2(r2−r +k )1+ 1 ( ) ; witch gives:
2 r +1
k

2( r − k ) + 1
2r + 1

 r
k 2( r − k ) + 1
 (− 1)
2r + 1
k =0

( )(∑ a
2 r +1
k

( )(a b )
2 r +1
k

i j

3

2 r − k +1 k

2 r − k +1

b

)(∑ a b

k 2 r − k +1

(a j bi )k 



)

The transposition of i& j; gives:
M 2;2r =

∑ ∑


2(r − k ) + 1
2r + 1
i < j k = 0
+

r

∑

2( r − k ) + 1
2r + 1
k =0
r

( )(a b )
2 r +1
k

( )(a b )
2 r +1
k

( )

i j

j i

2 r − k +1

2 r − k +1

(− a j bi )k +

(− ai b j )k 


 r 2(r − k ) + 1 2 r +1
2 r − k +1
(− a j bi )k +
M 2; 2 r = ∑ ∑
k (a i b j )
2r + 1
i < j  k =0
2r
 2(r − l ) + 1  2 r +1
2 r −l +1 
l
+ ∑−
 l (a j bi ) (− ai b j )

2r + 1 
l = r +1

2 r +1 2(r − k ) + 1
= ∑ ∑
2r + 1
i < j  k =0

( )

( )(a b )
2 r +1
k

2 r − k +1

i

j

let p = a j bi & q = −a i b j , we have:
M 2; 2 r =

∑ ∑

2 r +1 2(r − k ) + 1

i< j 
 k = 0 2r + 1

( )p
2 r +1
k

(− a b ) 
k

j i


q 


2 r − k +1 k



2 2 r +1 2 r +1 2 r − k +1 k 
2 r +1 2(r − k ) + 1 2 r +1 2 r − k +1 k
−
= ∑ ∑
p
q
(
)
∑ k ( k )p q 
k
2r + 1
r + 1 k =1
i < j  k =0
2(2r + 1) 2 r +1 2 r 2 r − k +1 k 
2 r +1
= ∑  ∑ ( 2 rk+1 ) p 2 r − k +1 q k −
∑ k ( k −1 )p q 
r + 1 k =1
i < j  k =0

2r


2 r +1
= ∑ ( p + q )
− 2∑ k ( 2lr ) p 2 r −l q l +1 
i< j 
l =1


=

=

∑ {( p + q )2r +1 − 2q( p + q )2r }

∑ {(a j bi − ai b j )2r (a j bi + ai b j )}
i< j

i< j

4

3. Application to polynomials.

Let P( x) =

xi = µ i

p

∏i =1 ( x − µ i ) and Tn = ∑ µ1
n

n

& yi = µ i

i =1

q

we have

n

( µ i is the module of µ i ) : For

∑ (−1) k  k Tkp+(m+1−k )qTkq+(m+1−k ) p ≥ 0
m

k =0

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;…
m = 1 → T2 p T2 q − T p2+q ≥ 0

m = 2 → T3 p T3q − 2T2 p + q T p + 2 q ≥ 0

m = 3 → T4 p T4 q − 4T3 p + q T p +3q + 2T22p + 2 q ≥ 0

m = 4 → T5 p T5q + 2T3 p + 2 q T2 p +3q − 3T4 p + q T p + 4q ≥ 0

m = 5 → T6 p T6 q + 15T4 p + 2 q T2 p + 4 q − 6T5 p + q T p +5q − 10T32p +3q ≥ 0

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, BabEzzouar, 16111 Algér, Algérie.

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