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J ournal of
Mathematical
I nequalities
Volume 3, Number 2 (2009), 237–242
ABOUT BERGSTRÖM’S INEQUALITY
OVIDIU T. P OP
(Communicated by P. Bullen)
Abstract. In this paper, we generalize identity (3), from where we obtain a rafinement of inequalities (1) and (2).
1. Introduction
The inequality from Theorem 1 is called in the literature Bergström’s inequality
(see [2], [3], [4], [6]).
T HEOREM 1. If xk ∈ R and ak > 0 , k ∈ {1, 2, . . . , n} , then
x21 x22
x2
(x1 + x2 + . . . + xn)2
+ + ...+ n ,
a1 a2
an
a1 + a2 + . . . + an
with equality if and only if
(1)
xn
x1 x2
= = ... =
.
a1 a2
an
The generalization of Bergström’s inequality is contained in the following theorem
(see [5]).
T HEOREM 2. If xk ∈ R and ak > 0 , k ∈ {1, 2, . . . , n} , then
(ai x j − ai x j )2
x21 x22
x2
(x1 + x2 + . . . + xn )2
.
+ + ...+ n + max
1i< jn ai a j (ai + a j )
a1 a2
an
a1 + a2 + . . . + an
(2)
By particularizations in Theorem 2, in paper [5] the refinements are obtained of
Cauchy-Schwarz’s inequality.
For the complex numbers, it is well-known the identity:
|z1 |2 |z2 |2 |z1 + z2 |2
|a1 z2 − a2z1 |2
+
−
=
a1
a2
a1 + a2
a1 a2 (a1 + a2)
(3)
Mathematics subject classification (2000): 26D15.
Keywords and phrases: Bergström’s inequality, Lagrange’s identity, multivariate inequalities refinements.
c
, Zagreb
Paper JMI-03-24
237
238
OVIDIU T. P OP
true for any z1 , z2 ∈ C and any a1 , a2 ∈ R\{0} such that a1 + a2 = 0 (see [6], page
315).
In the following, we note N = {1, 2, . . .}
2. Main results
In this section we start with the generalization of identity (3).
T HEOREM 3. If n ∈ N, n 2 , z1 , z2 , . . . , zn ∈ C and a1 , a2 , . . . , an ∈ R\{0} , such
that a1 + a2 + . . . + an = 0 , then
|z1 |2 |z2 |2
|zn |2 |z1 + z2 + . . . + zn |2
+
+ ...+
−
a1
a2
an
a1 + a2 + . . . + an
|ai z j − a j zi |2
1
=
.
∑
a1 + a2 + . . . + an 1i< jn
ai a j
(4)
Proof. We consider zk = xk + iyk , xk , yk ∈ R, k ∈ {1, 2, . . ., n} . Then the member
from the left from the (4) is equivalent with
x21 +y21 x22 +y22
x2 +y2 (x1 +x2 + . . . +xn )2 +(y1 +y2 + . . . +yn )2
+
+ ...+ n n −
a1
a2
an
a1 +a2 + . . . +an
1
=
a1 a2 · . . . · an (a1 + a2 + . . . + an)
· (a22 a3 a4 · . . . · an + a2 a23 a4 · . . . · an + . . . + a2a3 · . . . · an−1a2n )(x21 + y21 )
+(a21 a3 a4 · . . . ·an + a1 a23 a4 · . . . ·an + . . . +a1 a3 · . . . ·an−1 a2n )(x22 +y22 )+ . . .
+(a21 a2 a3 · . . . ·an−1 + a1a22 a3 · . . . ·an−1 + . . . +a1 a2 · . . . ·an−2 a2n−1 )(x2n +y2n )
− 2a1a2 · . . . · an (x1 x2 + x1x3 + . . . + x1 xn + . . . + xn−1xn )
+ (y1 y2 + y1 y3 + . . . + y1yn + . . . + yn−1yn ))
1
=
a3 a4 . . . an ((a1 x2 −a2 x1 )2 +(a1 y2 −a2 y1 )2 )
a1 a2 · . . . ·an (a1 +a2 + . . . +an )
+ a2 a4 a5 · . . . · an (a1 x3 − a3 x1 )2 + (a1 y3 − a3y1 )2 + . . .
+ a1 a2 · . . . · an−2 (an−1 xn − anxn−1 )2 + (an−1yn − an yn−1 )2
1
a3 a4 · . . . · an |a1 z2 − a2 z1 |2
=
a1 a2 · . . . · an (a1 + a2 + . . . + an)
+ a2 a4 a5 · . . . · an |a1 z3 − a3 z1 |2 + . . . + a1 a2 · . . . · an−2 |an−1zn − an zn−1 |2 ,
from where, the relation (4) results.
239
A BOUT B ERGSTR ÖM ’ S INEQUALITY
C OROLLARY 1. If n∈N, n2 , x1 , x2 , . . . , xn ∈R and a1 , a2 , . . . , an ∈R\{0} , such
that a1 + a2 + . . . + an = 0 , then
x21 x22
x2 (x1 + x2 + . . . + xn )2
+ + ...+ n −
a1 a2
an
a1 + a2 + . . . + an
(ai x j − a j xi )2
1
.
=
∑
a1 + a2 + . . . + an 1i< jn
ai a j
(5)
Proof. In the identity (4) we consider zk = xk ∈ R, k ∈ {1, 2, . . ., n} .
C OROLLARY 2. If n ∈ N, x 2 , xk , yk ∈ R, k ∈ {1, 2, . . . , n} , then
(x21 + x22 + . . . + x2n )(y21 + y22 + . . . + y2n )
(6)
2
− (x1 y1 + x2y2 + . . . + xnyn ) =
∑
2
(xi y j − x j yi ) .
1i< jn
Proof. Changing ak by y2k and xk by xk yk , k ∈ {1, 2, . . . , n} in identity (5), we
obtain identity (6).
R EMARK 1. The identity from Corollary 2 is called Lagrange’s identity.
C OROLLARY 3. If z1 , z2 , . . . , zn ∈ C and a1 , a2 , . . . , an ∈ (0, ∞), then
|z1 |2 |z2 |2
|zn |2
|z1 + z2 + . . . + zn |2
+
+ ...+
,
a1
a2
an
a1 + a2 + . . . + an
(7)
with equality if and only if ai z j = a j zi for any i, j ∈ {1, 2, . . . , n} .
Proof. The inequality (7) results immediately from Theorem 3.
T HEOREM 4. If n ∈ N, n 2 , z1 , z2 , . . . , zn ∈ C and a1 , a2 , . . . , an ∈ R\{0} such
that a1 + a2 + . . . + an > 0 , then
|ai z j −a j zi |2
|z1 |2 |z2 |2
|zn |2
1
+
+ ...+
,
∑
a1
a2
an
a1 +a2 + . . . +an 1i< jn
ai a j
(8)
with equality if and only if z1 + z2 + . . . + zn = 0 .
Proof. It results from Theorem 3.
C OROLLARY 4. If n ∈ N, n 2 and z1 , z2 , . . . , zn ∈ C, then
|z1 |2 + |z2 |2 + . . . + |zn |2 1
∑ |z j − zi|2 .
n 1i<
jn
(9)
240
1.
OVIDIU T. P OP
Proof. The inequality (9) results from Theorem 4 if we take a1 = a2 = . . . = an =
T HEOREM 5. If n ∈ N, n 2 , x1 , x2 , . . . , xn ∈ R and a1 , a2 , . . . , an ∈ (0, ∞), then
x21 x22
x2 (x1 + x2 + . . . + xn )2
+ + ...+ n −
a1 a2
an
a1 + a2 + . . . + an
(ai x j − a j xi )2
1
Ak,l +
,
∑
a1 + a2 + . . . + an 1i< jn
ai a j
(10)
i, j∈{k,l}
where
(ai x j − a j xi )2
(ak xl − al xk )2
=
, 1 k < l n.
1i< jn ai a j (ai + a j )
ak al (ak + al )
Ak,l = max
Proof. We have that
T=
n
∑
m=1
m∈{k,l}
=
(am xl − al xm )2 (am xk − ak xm )2
+
am al
am ak
⎛
2 ⎞
al ak
al ak
x
−
x
x
−
a
x
a
l k
am m
⎜ k l am m
⎟
+
⎝
⎠
a a2
a a2
m
∑
l k
m=1
m∈{k,l}
k l
am
am
and applying the inequality (1) for n = 2 , we obtain that
n
T
(ak xl − al xk )2
∑
m=1
m∈{k,l}
ak al (ak +al )
am
n
=
∑
m=1
m∈{k,l}
am (ak xl − al xk )2
.
ak al (ak + al )
Taking the inequality above into account, we have that
(ai x j −a j xi )2
1
∑
a1 +a2 + . . . +an 1i<
ai a j
jn
1
(ak xl −al xk )2
=
+
a1 +a2 + . . . +an
ak al
+
1
a1 +a2 + . . . +an
1
a1 +a2 + . . . +an
+
∑
1i< jn
i, j∈{k,l}
∑
∑
m=1
m∈{k,l}
(am xl −al xm )2 (am xk −ak xm )2
+
am al
a m xk
(ai x j −a j xi )2
ai a j
(ak xl −al xk )2
+
ak al
1
a1 + a2 + . . . + an
n
1i< jn
i, j∈{k,l}
n
∑
m=1
m∈{k,l}
am (ak xl − al xk
ak al (ak + al )
(ai x j − a j xi )2
,
ai a j
from where and taking identity (5) into account, the inequality (10) results.
241
A BOUT B ERGSTR ÖM ’ S INEQUALITY
R EMARK 2. From Theorem 5 the inequalities from (1) and (2) results.
T HEOREM 6. If n ∈ N, n 2 , xi , yi ∈ R, i ∈ {1, 2, . . ., 3} , then
n
∑
i=1
n
∑
x2i
i=1
y2i
n
∑
i=1
∑ xi yi
−
y2i
2
n
(11)
i=1
(xk yl − xl yk )2
+ ∑ (xi y j − x j yi )2
y2k + y2l
1i< jn
i, j∈{k,l}
if yi = 0 for any i ∈ {1, 2, . . . , n} ,
n
∑
i=1
n
∑
x2i
i=1
y2i
n
i=1
2
∑ xi yi
−
∑ x2i
n
(12)
i=1
(x p yq − xq y p )2
+ ∑ (xi y j − x j yi )2
x2p + x2q
1i< jn
i, j∈{p,q}
if xi = 0 for any i ∈ {1, 2, . . ., } , where k, l, p, q ∈ {1, 2, . . . , n} , k = l , p = q such that
(xi y j − x j yi )2
(x y −x y )2
max i yj2 +yj2 i is obtained for i = k and j = l , max
is obtained
1i< jn
1i< jn
i
j
x2i + x2j
for i = p and j = q .
Proof. In Theorem 5 we change ai by y2i and xi with xi yi , i ∈ {1, 2, . . . , n} and
we obtain the inequality (11), respectively xi by yi , i ∈ {1, 2, . . ., n} in inequality (11)
and then we obtain inequality (12).
C OROLLARY 5. If n ∈ N, n 2 , xi , yi ∈ R\{0} , i ∈ {1, 2, . . ., n} , then the inequalities
n
∑
i=1
and
n
x2i
∑ x2i
i=1
n
∑
i=1
n
y2i
i=1
2
∑ xi yi
−
−
n
∑ xi yi
i=1
n
∑
i=1
∑ y2i
n
i=1
2
n
y2i
∑ xi
i=1
2
(xi y j − x j yi )2
y2i + y2j
(13)
(xi y j − x j yi )2
x2i + y2j
(14)
hold for any i, j ∈ {1, 2, . . ., n} .
Proof. It results from Theorem 6.
R EMARK 3. Using the inequalities from Corollary 5, we obtain some inequalities
from paper [5].
242
OVIDIU T. P OP
REFERENCES
[1] G. H ARDY, J. E. L ITTLEWOOD AND G. P ÓLYA, Inequalities, Cambridge, 1934.
[2] E. F. B ECHENCBACH AND R. B ELLMAN, Inequalities, Springer, Berlin, Göttingen and Heidelberg,
1961.
[3] R. B ELLMAN, Notes on Matrix Theory - IV (An Inequality Due to Bergström), Amer. Math. Montly,
Vol. 62 (1955), 172–173
[4] H. B ERGSTR ÖM, A triangle inequality for matrices, Den Elfte Skandinaviske Matematikerkongress,
1949, Trodheim, Johan Grundt Tanums Forlag, Oslo, 1952, 264–267.
[5] D. M ĂRGHIDANU , J. L. D ÍAZ -BARRERO AND S. R ĂDULESCU, New refinements of some classical
inequalities, to appear in Math. Ineq.& Appl.
[6] D. S. M ITRINOVI Ć, Analytic Inequalities, Springer-Verlag, Berlin, 1970.
(Received October 22, 2008)
Journal of Mathematical Inequalities
www.ele-math.com
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Ovidiu T. Pop
National College “Mihai Eminescu”
5 Mihai Eminescu Street
Satu Mare 440014
Romania
e-mail: [email protected]