Volume 8 (2007), Issue 3, Article 78, 5 pp.
INEQUALITIES BETWEEN THE SUM OF SQUARES AND THE EXPONENTIAL
OF SUM OF A NONNEGATIVE SEQUENCE
FENG QI
R ESEARCH I NSTITUTE OF M ATHEMATICAL I NEQUALITY T HEORY
H ENAN P OLYTECHNIC U NIVERSITY
J IAOZUO C ITY, H ENAN P ROVINCE , 454010, C HINA
qifeng@hpu.edu.cn
URL: http://rgmia.vu.edu.au/qi.html
Received 11 April, 2007; accepted 08 September, 2007
Communicated by A. Sofo
A BSTRACT. Using a standard argument, the following inequality between the sum of squares
and the exponential of sum of a nonnegative sequence is established:
!
n
n
X
e2 X 2
x ≤ exp
xi ,
4 i=1 i
i=1
where n ≥ 2, xi ≥ 0 for 1 ≤ i ≤ n, and the constant
e2
4
is the best possible.
Key words and phrases: Inequality, Sum of square, Exponential of sum, Nonnegative sequence, Critical point, Extremal point,
Open problem.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
In the 2004 Master Graduate Admission Examination of Mathematical Analysis of the Beijing Institute of Technology, the following inequality, which was brought up by one of the
author’s students, was asked to be shown: For (x, y) ∈ [0, ∞) × [0, ∞), show
x2 + y 2
≤ exp(x + y − 2).
4
The aim of this paper is to give a generalization of inequality (1.1).
For our own convenience, we introduce the following notations:
(1.1)
(1.2)
[0, ∞)n , [0, ∞) × [0, ∞) × · · · × [0, ∞)
|
{z
}
n times
and
(1.3)
(0, ∞)n , (0, ∞) × (0, ∞) × · · · × (0, ∞)
|
{z
}
n times
for n ∈ N, where N denotes the set of all positive integers.
112-07
2
F. Q I
The main results of this paper are the following theorems.
Theorem 1.1. For (x1 , x2 , . . . , xn ) ∈ [0, ∞)n and n ≥ 2, inequality
!
n
n
X
e2 X 2
(1.4)
x ≤ exp
xi
4 i=1 i
i=1
is valid. Equality in (1.4) holds if xi = 2 for some given 1 ≤ i ≤ n and xj = 0 for all 1 ≤ j ≤ n
2
with j 6= i. So, the constant e4 in (1.4) is the best possible.
P∞
Theorem 1.2. Let {xi }∞
i=1 be a nonnegative sequence such that
i=1 xi < ∞. Then
!
∞
∞
X
e2 X 2
(1.5)
x ≤ exp
xi .
4 i=1 i
i=1
Equality in (1.5) holds if xi = 2 for some given i ∈ N and xj = 0 for all j ∈ N with j 6= i. So,
2
the constant e4 in (1.5) is the best possible.
Remark 1.3. Taking n = 2 and (x1 , x2 ) = (x, y) in (1.4) easily leads to inequality (1.1).
Taking xi = x and xj = y for some given i, j ∈ N and xk = 0 for all k ∈ N with k 6= i and
k 6= j in inequality (1.5) also clearly leads to inequality (1.1).
Remark 1.4. Inequality (1.4) can be rewritten as
n
n
e2 X 2 Y xi
x ≤
e
4 i=1 i
i=1
(1.6)
or
e2
kxk22 ≤ exp kxk1 ,
4
(1.7)
where x = (x1 , . . . , xn ) and k · kp denotes the p-norm.
Remark 1.5. Inequality (1.5) can be rewritten as
∞
∞
e2 X 2 Y xi
x ≤
e
4 i=1 i
i=1
(1.8)
which is equivalent to inequality (1.7) for x = (x1 , x2 , . . . ) ∈ [0, ∞)∞ .
Remark 1.6. Taking xi =
(1.9)
Taking xi =
(1.10)
1
i
for i ∈ N in (1.4) and rearranging gives
!
n
n
X
X
1
1
2 − 2 ln 2 + ln
≤
.
2
i
i
i=1
i=1
1
is
for i ∈ N and s > 1 in (1.5) and rearranging gives
!
∞
∞
X
X
1
1
2 − 2 ln 2 + ln
= 2 − 2 ln 2 + ln[ζ(2s)] ≤
= ζ(s),
2s
i
is
i=1
i=1
where ζ denotes the well known Riemann Zeta function.
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 78, 5 pp.
http://jipam.vu.edu.au/
I NEQUALITIES B ETWEEN S UM OF S QUARES AND E XPONENTIAL OF S UM
3
2. P ROOFS OF T HEOREMS
Now we are in a position to prove our theorems.
Proof of Theorem 1.1. Let
f (x1 , x2 , . . . , xn ) = ln
(2.1)
n
X
!
x2i
i=1
−
n
X
xi
i=1
for (x1 , x2 , . . . , xn ) ∈ [0, ∞)n \ {(0, 0, . . . , 0)}. Simple calculation results in
2xk
∂f (x1 , x2 , . . . , xn )
= Pn 2 − 1,
∂xk
xi
i=1
Pn
2
2
2
i6=k xi − xk
∂ 2 f (x1 , x2 , . . . , xn )
=
,
P
2
∂x2k
( ni=1 x2i )
(2.2)
(2.3)
∂ 2 f (x1 , x2 , . . . , xn )
4x` xm
= − Pn 2 2 ,
∂x` ∂xm
( i=1 xi )
(2.4)
where 1 ≤ k, `, m ≤ n and ` 6= m. The system of equations
∂f (x1 , x2 , . . . , xn )
= 0 for 1 ≤ k ≤ n,
∂xk
(2.5)
which is equivalent to
X
(2.6)
x2i + (xk − 1)2 = 1 for 1 ≤ k ≤ n,
i6=k
has a unique nonzero solution xi = n2 for 1 ≤ i ≤ n. Thus, the point n2 , n2 , . . . , n2 is a unique
critical point of the function f (x1 , x2 , . . . , xn ), which is located in the interior of [0, ∞)n \
{(0, 0, . . . , 0)}.
Straightforward computation gives us
∂ 2 f n2 , n2 , . . . , n2
n−2
(2.7)
=
,
2
∂xk
2
∂ 2 f n2 , n2 , . . . , n2
= −1,
∂x` ∂xm
(2.8)
2 2 2
2 2
2
2 2
2 2
2
∂ f , , . . . , 2
∂
f
,
,
.
.
.
,
∂
f
,
,
.
.
.
,
n
n
n
n
n
n
n
n
n
···
∂x21
∂x1 ∂x2
∂x1 ∂xi
2 2 2
2
2
2
2
2
2
2
2
2
∂ f , , . . . ,
∂ f n, n, . . . , n
∂ f n , n , . . . , n n n
n
···
2
Di = ∂x2 ∂x1
∂x2
∂x2 ∂xi
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 2 2
2 2
2
2 2
2 2
2
∂ f n , n , . . . , n2
∂ f n, n, . . . , n
∂ f n, n, . . . , n ···
∂xi ∂x1
∂xi ∂x2
∂x2i
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 78, 5 pp.
http://jipam.vu.edu.au/
4
F. Q I
n − 2
−1
·
·
·
−1
2
n
−
2
−1
···
−1 =
2
. . . . . . . . . . . . . . . . . . . . . . . . . .
n − 2 −1 · · ·
−1
2
ih n − 2
ii−1
hn − 2
=
+ (i − 1)(−1)
− (−1)
2
2
n
n i−1
=
−i
.
2
2
Since

> 0, if i < n2 ,



(2.9)
Di = 0, if i = n2 ,



< 0, if i > n2 ,
it is affirmed that the critical point n2 , n2 , . . . , n2 located in the interior of [0, ∞)n \{(0, 0, . . . , 0)}
is not an extremal point of the function f (x1 , x2 , . . . , xn ).
i
n−i−1
The boundary of [0, ∞)n \ {(0, 0, . . . , 0)} is ∪n−1
.
i=0 [0, ∞) × {0} × [0, ∞)
n−1
On the set [0, ∞)
× {0} \ {(0, 0, . . . , 0)}, it is concluded that
! n−1
n−1
X
X
(2.10)
f (x1 , . . . , xn−1 , 0) = ln
x2k −
xk .
k=1
k=1
By the same standard argument as above, it is deduced that the unique critical point, located
2
2
in the interior of [0, ∞)n−1 × {0} \ {(0, 0, . . . , 0)}, of f (x1 , . . . , xn−1 , 0) is n−1
, . . . , n−1
,0
which is not an extremal point of f (x1 , . . . , xn−1 , 0).
By induction, in the interior of the set [0, ∞)i × {0} × · · · × {0} \{(0, 0, . . . , 0)} for 2 ≤ i ≤
|
{z
}
n − i times
n, there is no extremal point of f (x1 , . . . , xi , 0, . . . , 0).
On the set (0, ∞) × {0} × · · · × {0}, it is easy to obtain that the function
|
{z
}
n − 1 times
f (x1 , 0, . . . , 0) = 2 ln x1 − x1
has a maximal point x1 = 2 and the maximal value equals f (2, 0, . . . , 0) = 2 ln 2 − 2.
Considering that the function f (x1 , x2 , . . . , xn ) is symmetric with respect to all permutations
of the n variables xi for 1 ≤ i ≤ n and by induction, we obtain the following conclusion: The
maximal value of the function f (x1 , . . . , xn ) on the set [0, ∞)n \ {(0, 0, . . . , 0)} is 2 ln 2 − 2.
Therefore, it follows that
!
n
n
X
X
2
(2.11)
f (x1 , x2 , . . . , xn ) = ln
xi −
xi ≤ 2 ln 2 − 2,
i=1
i=1
which is equivalent to inequality (1.4), on the set [0, ∞)n \ {(0, 0, . . . , 0)}.
It is clear that inequality (1.4) holds also at the point (0, . . . , 0). Hence, the proof of Theorem
1.1 is complete.
Proof of Theorem 1.2. This can be concluded by letting n → ∞ in Theorem 1.1.
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 78, 5 pp.
http://jipam.vu.edu.au/
I NEQUALITIES B ETWEEN S UM OF S QUARES AND E XPONENTIAL OF S UM
5
3. O PEN P ROBLEMS
Finally, the following problems can be proposed.
Open Problem 1. For (x1 , x2 , . . . , xn ) ∈ [0, ∞)n and n ≥ 2, determine the best possible
constants αn , λn ∈ R and 0 < βn , µn < ∞ such that
!
n
n
n
X
X
X
(3.1)
βn
xαi n ≤ exp
x i ≤ µn
xλi n .
i=1
i=1
i=1
Open Problem 2. What is the integral analogue of the double inequality (3.1)?
Open Problem 3. Can one find applications and practical meanings in mathematics for inequality (3.1) and its integral analogues?
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 78, 5 pp.
http://jipam.vu.edu.au/