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INEQUALITIES INVOLVING THE MEAN AND
THE STANDARD DEVIATION OF NONNEGATIVE
REAL NUMBERS
OSCAR ROJO
Received 22 December 2005; Revised 18 August 2006; Accepted 21 Septe mber 2006
Let m(y)
=

n
j
=1
y
j
/n and s(y) =

m(y
2
) −m
2
(y) be the mean and the standard deviation
of the components of the vector y
= (y
1
, y
2
, , y
n−1
, y
n
), where y
q

= (y
q
1
, y
q
2
, , y
q
n
−1
, y
q
n
)
with q a positive integer. Here, we prove that if y
≥ 0,thenm(y
2
p
)+(1/
√
n −1)s(y
2
p
) ≤

m(y
2
p+1
)+(1/
√

n −1)s(y
2
p+1
)forp = 0,1,2, The equality holds if and only if
the (n
− 1) largest components of y are equal. It follows that (l
2
p
(y))
∞
p=0
, l
2
p
(y) =
(m(y
2
p
)+(1/
√
n −1)s(y
2
p
))
2
−p
, is a strictly increasing sequence converging to y
1
,the
largest component of y,exceptifthe(n

−1) largest components of y are equal. In this
case, l
2
p
(y) = y
1
for all p.
Copyright © 2006 Oscar Rojo. This is an op en access article distributed under the Cre-
ative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
1. Introduction
Let
m(x)
=

n
j
=1
x
j
n
, s(x)
=

m

x
2

−

m
2
(x) (1.1)
be the mean and the standard deviation of the components of x
= (x
1
,x
2
, , x
n−1
,x
n
),
where x
q
= (x
q
1
,x
q
2
, , x
q
n
−1
,x
q
n
) for a positive integer q.
The following theorem is due to Wolkowicz and Styan [3, Theorem 2.1.].

Theorem 1.1. Let
x
1
≥ x
2
≥···≥x
n−1
≥ x
n
. (1.2)
Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2006, Article ID 43465, Pages 1–15
DOI 10.1155/JIA/2006/43465
2 Inequalities on the mean and standard deviation
Then
m(x)+
1
√
n −1
s(x)
≤ x
1
, (1.3)
x
1
≤ m(x)+
√
n −1s(x). (1.4)
Equality holds in ( 1.3)ifandonlyifx

1
= x
2
=···=x
n−1
. Equality holds in (1.4)ifandonly
if x
2
= x
3
=···=x
n
.
Let x
1
,x
2
, ,x
n−1
,x
n
be complex numbers such that x
1
is a positive real number and
x
1
≥


x

2


≥···≥


x
n−1


≥


x
n


. (1.5)
Then,
x
p
1
≥


x
2


p

≥···≥


x
n−1


p
≥


x
n


p
(1.6)
for any positive integer p.WeapplyTheorem 1.1 to (1.6)toobtain
m

|
x|
p

+
1
√
n −1
s


|
x|
p

≤
x
p
1
,
x
p
1
≤ m

|
x|
p

+
√
n −1s

|
x|
p

,
(1.7)
where
|x|=(|x

1
|,|x
2
|, , |x
n−1
|,|x
n
|).
Then,
l
p
(x) =

m

|
x|
p

+
1
√
n −1
s

|
x|
p



1/p
(1.8)
isasequenceoflowerboundsforx
1
and
u
p
(x) =

m

|
x|
p

+
√
n −1s

|
x|
p

1/p
(1.9)
isasequenceofupperboundsforx
1
.
We recall that the p-norm and the infinity-norm of a vector x
= (x

1
,x
2
, , x
n
)are
x
p
=

n

i=1


x
i


p

1/p
,1≤ p<∞,
x
∞
= max
i


x

i


.
(1.10)
It is well known that lim
p→∞
x
p
=x
∞
.
Oscar Rojo 3
Then,
l
p
(x) =
⎛
⎜
⎝

x
p
p
n
+
1

n(n −1)





x
2p
2p
−

x
2p
p
n
⎞
⎟
⎠
1/p
,
u
p
(x) =
⎛
⎜
⎝

x
p
p
n
+


n −1
n




x
2p
2p
−

x
2p
p
n
⎞
⎟
⎠
1/p
.
(1.11)
In [2, Theorem 11], we proved that if y
1
≥ y
2
≥ y
3
≥···≥y
n
≥ 0, then

m

y
2
p

+
√
n −1s

y
2
p

≥

m

y
2
p+1

+
√
n −1s

y
2
p+1


(1.12)
for p
= 0,1,2, The equality holds if and only if y
2
= y
3
=···=y
n
. Using this inequal-
ity, we proved in [2, Theorems 14 and 15] that if y
2
= y
3
=···=y
n
,thenu
p
(y) = y
1
for all p,andify
i
<y
j
for some 2 ≤ j<i≤ n,then(u
2
p
(y))
∞
p=0
is a strictly decreasing

sequence converging to y
1
.
The main purpose of this paper is to prove that if y
1
≥ y
2
≥ y
3
≥···≥y
n
≥ 0, then
m

y
2
p

+
1
√
n −1
s

y
2
p

≤


m

y
2
p+1

+
1
√
n −1
s

y
2
p+1

(1.13)
for p
= 0,1,2, The equality holds if and only if y
1
= y
2
=···=y
n−1
. Using this in-
equality, we prove that if y
1
= y
2
=···=y

n−1
,thenu
p
(y) = y
1
for all p,andify
i
<y
j
for
some 1
≤ j<i≤ n −1, then (l
2
p
(y))
∞
p=0
is a strictly increasing sequence converging to y
1
.
2. New inequalities involving m(x) and s(x)
Theorem 2.1. Let x
= (x
1
,x
2
, , x
n−1
,x
n

) be a vector of complex numbers s uch that x
1
is a
positive real number and
x
1
≥


x
2


≥···≥


x
n−1


≥


x
n


. (2.1)
The sequence (l
p

(x))
∞
p=1
converges to x
1
.
Proof. From (1.11),
l
p
(x) ≥

x
p
p
√
n
∀p. (2.2)
Then, 0
≤|l
p
(x) −x
1
|=x
1
−l
p
(x) ≤ x
1
−x
p

/
p
√
n for all p.Sincelim
p→∞
 x 
p
= x
1
and lim
p→∞
p
√
n=1, it follows that the sequence (l
p
(x)) converges and lim
p→∞
l
p
(x) =x
1
.

We introduce the following notations:
(i) e
=(1,1, ,1),
(ii) Ᏸ
= R
n
−{λe :λ ∈ R},

(iii) Ꮿ
={x =(x
1
,x
2
, , x
n
):0≤ x
k
≤ 1, k = 1,2, ,n},
4 Inequalities on the mean and standard deviation
(iv) Ᏹ
={x =(1,x
2
, , x
n
):0≤ x
n
≤ x
n−1
≤···≤x
2
≤ 1},
(v)
x,y=

n
k
=1
x

k
y
k
for x,y ∈ R
n
,
(vi)
∇g(x) =(∂
1
g(x),∂
2
g(x), ,∂
n
g(x)) denotes the gradient of a differentiable func-
tion g at the point x,where∂
k
g(x) is the partial derivative of g with respect to x
k
,
evaluated at x.
Clearly, if x
∈ Ᏹ,thenx
q
∈ Ᏹ with q a positive integer.
Let v
1
,v
2
, ,v
n

be the points
v
1
= (1,0, ,0),
v
2
= (1,1,0, ,0),
v
3
= (1,1,1,0, ,0),
.
.
.
v
n−2
= (1,1, ,1,0,0),
v
n−1
= (1,1, ,1,1,0),
v
n
= (1,1, ,1,1) =e.
(2.3)
Observe that v
1
,v
2
, , v
n
lie in Ᏹ.Foranyx =(1,x

2
,x
3
, ,x
n−1
,x
n
) ∈Ᏹ,wehave
x
=

1 −x
2

v
1
+

x
2
−x
3

v
2
+

x
3
−x

4

v
3
+ ···+

x
n−2
−x
n−1

v
n−2
+

x
n−1
−x
n

v
n−1
+ x
n
v
n
.
(2.4)
Therefore, Ᏹ is a convex set. We define the function
f (x)

= m(x)+
1
√
n −1
s(x), (2.5)
where x
= (x
1
,x
2
, , x
n
) ∈R
n
.Weobservethat
ns
2
(x) =
n

k=1
x
2
k
−


n
j
=1

x
j

2
n
=
n

k=1

x
k
−m(x)

2
=


x−m(x)e


2
2
.
(2.6)
Then,
f (x)
= m(x)+
1


n(n −1)


x−m(x)e


2
=

n
j
=1
x
j
n
+
1

n(n −1)






n

k=1
x
2

k
−


n
j
=1
x
j

2
n
.
(2.7)
Next, we give properties of f . Some of the proofs are similar to those in [2].
Oscar Rojo 5
Lemma 2.2. The function f has continuous first partial derivatives on Ᏸ,andforx
=
(x
1
,x
2
, , x
n
) ∈Ᏸ and 1 ≤k ≤ n,
∂
k
f (x) =
1
n

+
1
n(n −1)
x
k
−m(x)
f (x) −m(x)
, (2.8)
n

k=1
∂
k
f (x) =1, (2.9)

∇
f (x),x

=
f (x). (2.10)
Proof. From (2.7), it is clear that f is differentiable at every point x
= m(x )e,andfor
1
≤ k ≤n,
∂
k
f (x) =
1
n
+

1

n(n −1)
x
k
−

n
j
=1
x
j
/n


n
i
=1
x
2
i
−


n
j
=1
x
j


2
/n
=
1
n
+
1
n(n −1)
x
k
−m(x)
f (x) −m(x)
,
(2.11)
which is a continuous function on Ᏸ. Then,

n
k
=1
∂
k
f (x) =1. Finally,

∇
f (x),x

=
n

k=1

x
k
∂
k
f (x)
=

n
k
=1
x
k
n
+
1
n(n −1)

n
k
=1
x
2
k
−m(x)

n
k
=1
x
k

f (x) −m(x)
= m(x)+
1

n(n −1)


x −a(x)e


2
= f (x).
(2.12)
This completes the proof.

Lemma 2.3. The function f is convex on Ꮿ.Moreprecisely,forx,y ∈ Ꮿ and t ∈[0,1],
f

(1 −t)x + ty

≤
(1 −t) f (x)+tf(y) (2.13)
with equality if and only if
x
−m(x)e =α

y −m(y)e

(2.14)
for some α

≥ 0.
Proof. Clearly Ꮿ is a convex set. Let x, y
∈ Ꮿ and t ∈[0,1]. Then,
f

(1 −t)x + ty

=
m

(1 −t)x + ty

+
1

n(n −1)


(1 −t)x + ty −m

(1 −t)x + ty

e


2
=(1−t)m(x)+tm(y )+
1

n(n−1)



(1−t)

x−m(x)e

+t

y−m(y)e



2
.
(2.15)
6 Inequalities on the mean and standard deviation
Moreover ,


(1 −t)

x −m(x)e

+ t

y −m(y)e



2

2
= (1 −t)
2


x −m(x)e


2
2
+2(1−t)t

x −m(x)e,y −m(y)e

+ t
2


y −m(y)e


2
2
.
(2.16)
We recall the Cauchy-Schwarz inequality to obtain

x −m(x)e,y −m(y)e

≤



x −m(x)e


2


y −m(y)e


2
(2.17)
with equality if and only if (2.14)holds.Thus,


(1 −t)

x −m(x)e

+ t

y −m(y)e



2
≤ (1−t)



x −m(x)e


2
+ t


y −m(y)e


2
(2.18)
with equality if and only if (2.14)holds.Finally,from(2.15)and(2.18), the lemma fol-
lows.

Lemma 2.4. For x,y ∈ Ᏹ −{e},
f (x)
≥

∇f (y),x

(2.19)
with equality if and only if (2.14) holds for some α>0.
Proof. Ᏹ isaconvexsubsetofᏯ and f isaconvexfunctiononᏱ.Moreover, f isadiffer-
entiable function on Ᏹ
−{e}.Letx,y ∈ Ᏹ −{e}.Forallt ∈[0, 1],
f

tx+(1 −t)y


≤
tf(x)+(1−t) f (y). (2.20)
Thus, for 0 <t
≤ 1,
f

y + t(x −y)

−
f (y)
t
≤ f (x) − f (y). (2.21)
Letting t
→ 0
+
yields
lim
t→0
+
f

y + t(x −y)

−
f (y)
t
=

∇
f (y),x −y


≤
f (x) − f (y). (2.22)
Hence,
f (x)
− f (y) ≥

∇f (y),x

−

∇
f (y),y

. (2.23)
Now, we use the fact that
∇f (y),y=f (y)toconcludethat
f (x)
≥

∇f (y),x

. (2.24)
The equality in all the above inequalities holds if and only if x
−a(x)e =α(y −m(y)e)for
some α
≥ 0. 
Oscar Rojo 7
Corollar y 2.5. For x
∈ Ᏹ −{e},

f (x)
≥

∇f

x
2

,x

, (2.25)
where
∇f (x
2
) is the gradient of f with respect to x evaluated at x
2
. The equality in (2.25)
holds if and only if x is one of the following convex combinations:
x
i
(t) =te+(1 −t)v
i
, i =1, 2, ,n −1, some t ∈[0,1). (2.26)
Proof. Let x
= (1,x
2
,x
3
, , x
m

) ∈Ᏹ −{e}.Then,x
2
∈ Ᏹ −{e}. Using Lemma 2.4,weob-
tain
f (x)
≥

∇f

x
2

,x

(2.27)
with equality if and only if
x
−m(x)e =α

x
2
−m

x
2

e

(2.28)
for some α

≥ 0. Thus, we have proved (2.25). In order to complete the proof, we observe
that condition (2.28)isequivalentto
x
−αx
2
= m

x −αx
2

e (2.29)
for some α
≥ 0. Since x
1
= 1, (2.29)isequivalentto
1
−α =x
2
−αx
2
2
= x
3
−αx
2
3
=···=x
n
−αx
2

n
(2.30)
for some α
≥ 0. Hence, (2.28)isequivalentto(2.30).
Suppose that (2.30)istrue.Ifα
= 0, then 1 = x
2
=···=x
n
. This is a contradiction
because x
= e,thusα>0.
If x
2
= 0, then x
3
= x
4
=···=x
n
= 0, and thus x =v
1
.Let0<x
2
< 1. Suppose x
3
<x
2
.
From (2.30),

1
−x
2
= α

1+x
2

1 −x
2

,
x
2
−x
3
= α

x
2
+ x
3

x
2
−x
3

.
(2.31)

From these equations, we obtain x
3
= 1, which is a contradiction. Hence, 0 <x
2
< 1im-
plies x
3
= x
2
.Now,ifx
4
<x
3
,fromx
2
= x
3
and the equations
1
−x
2
= α

1+x
2

1 −x
2

,

x
3
−x
4
= α

x
3
+ x
4

x
3
−x
4

,
(2.32)
we obtain x
4
= 1, which is a contradiction. Hence, x
4
= x
3
if 0 <x
2
< 1. We continue in
this fashion to conclude that x
n
= x

n−1
=···=x
3
= x
2
.Wehaveprovedthatx
1
= 1and
0
≤ x
2
< 1implythatx =(1,t, ,t) =te +(1−t)v
1
for some t ∈ [0,1). Let x
2
= 1.
8 Inequalities on the mean and standard deviation
If x
3
= 0, then x
4
= x
5
=···=x
m
= 0, and thus x =v
2
.Let0<x
3
< 1andx

4
<x
3
.
From (2.30),
1
−x
3
= α

1+x
3

1 −x
3

,
x
3
−x
4
= α

x
3
+ x
4

x
3

−x
4

.
(2.33)
From these equations, we obtain x
4
= 1, which is a contradiction. Hence, 0 <x
3
< 1im-
plies x
4
= x
3
.Now,ifx
5
<x
4
,fromx
3
= x
4
and the equations
1 −x
3
= α

1+x
3


1 −x
3

,
x
4
−x
5
= α

x
4
+ x
5

x
4
−x
5

,
(2.34)
we obtain x
5
= 1, which is a contradiction. Therefore, x
5
= x
4
. We continue in this fashion
to get x

n
= x
n−1
=···=x
3
.Thus,x
1
= x
2
= 1, and 0 ≤x
3
< 1 implies that x = (1,1,t, ,t)
= te+(1 −t)v
2
for some t ∈ [0,1).
For 3
≤ k ≤ n −2, arguing as above, it can be proved that x
1
= x
2
=···=x
k
= 1and
0
≤ x
k+1
< 1 implies that x = (1, ,1,t, ,t) = te+(1 −t)v
k
.Finally,forx
1

= x
2
=···=
x
n−1
= 1and0≤x
n
< 1, we have x =te + v
n−1
.
Conversely, if x is any of the convex combinations in (2.26), then (2.30)holdsby
choosing α
= 1/(1 + t). 
Let us define the following optimization problem.
Problem 2.6. Let
F :
R
n
−→ R (2.35)
be given by
F(x)
= f

x
2

−

f (x)


2
. (2.36)
We want to find m in
x∈Ᏹ
F(x). That is, find
minF(x) (2.37)
subject to the constraints
h
1
(x) =x
1
−1 =0,
h
i
(x) =x
i
−x
i−1
≤ 0, 2 ≤ i ≤ n,
h
n+1
(x) =−x
n
≤ 0.
(2.38)
Lemma 2.7. (1) If x
∈ Ᏹ −{e}, then

n
k

=1
∂
k
F(x) ≤0 with equality if and only if x is one of
the convex combinations x
k
(t) in (2.26).
(2) If x
= x
N
(t) with 1 ≤N ≤ n −2, then
∂
1
F(x) =···=∂
N
F(x) > 0, (2.39)
∂
N+1
F(x) =···=∂
n
F(x) < 0. (2.40)
Oscar Rojo 9
Proof. (1) The function F has continuous first partial derivatives on Ᏸ,andforx
∈ Ᏸ
and 1
≤ k ≤n,
∂
k
F(x) =2x
k

∂
k
f (x
2
) −2 f (x)∂
k
f (x). (2.41)
By (2.9),
n

k=1
∂
k
F(x) =2
n

k=1
x
k
∂
k
f

x
2

−
2 f (x)
n


k=1
∂
k
f (x)
= 2

∇
f

x
2

,x

−
2 f (x).
(2.42)
It follows from Corollary 2.5 that

n
k
=1
∂
k
F(x) ≤ 0 with equality if and only if x
i
= te+
(1
−t)v
i

, i = 1, ,n−1.
(2) Let x
= x
N
(t)with1≤ N ≤ n −2fixed.Then,x = te+(1 −t)v
N
,somet ∈ [0,1).
Thus, x
1
= x
2
=···=x
N
= 1, x
N+1
= x
N+2
=···=x
n
= t.FromTheorem 1.1, f (x) < 1.
Moreover ,
f (x)
−m(x) =

1
n(n −1)



N +(n −N)t

2
−

N +(n −N)t

2
n
=

1
n(n −1)

nN + n(n −N)t
2
−N
2
−2N(n −N)t −(n −N)
2
t
2
n
=
1
n
√
n −1

N(n −N)(1−t).
(2.43)
Replacing this result in (2.8), we obtain

∂
1
f (x) =∂
2
f (x) =···=∂
N
f (x)
=
1
n
+
1
n(n −1)
1
−m(x)
f (x) −m(x)
=
1
n
+
1
√
n −1
1
−

N +(n −N)t

/n


N(n −N)(1−t)
=
1
n
+
1
√
n −1n
√
n −N
√
N
> 0.
(2.44)
Similarly,
f

x
2

−
m

x
2

=
1
n
√

n −1

N(n −N)

1 −t
2

,
∂
1
f

x
2

=
∂
2
f

x
2

=···=
∂
N
f

x
2


=
1
n
+
1
n
√
n −1
√
n −N
√
N
> 0.
(2.45)
10 Inequalities on the mean and standard deviation
Therefore,
∂
1
F(x) =∂
2
F(x) =···=∂
N
F(x)
= 2∂
1
f

x
2


−
2 f (x)∂
1
f (x) =2

1 − f (x)

∂
1
f (x) > 0.
(2.46)
We have thu s proved (2.39). We easily see that
∂
N+1
F(x) =∂
N+2
F(x) =···=∂
n
F(x). (2.47)
We have

n
k
=1
∂
k
F(x) =0. Hence,
n


k=N+1
∂
k
F(x) =(n −N)∂
N+1
F(x) =−
N

k=1
∂
k
F(x) < 0. (2.48)
Thus, (2.40)follows.

We recall the following necessary condition for the existence of a minimum in nonlin-
ear programming.
Theorem 2.8 (see [1, Theorem 9.2-4(1)]). Let J : Ω
⊆ V → R be a function defined over
an open, convex subset Ω of a Hilbert space V and let
U
=

v ∈Ω : ϕ
i
(v) ≤0, 1 ≤i ≤m

(2.49)
be a subset of Ω, the constraints ϕ
i
: Ω → R, 1 ≤ i ≤ m, being assumed to be convex. Let

u
∈ U be a point at which the functions ϕ
i
, 1 ≤i ≤m,andJ are differentiable. If the function
J has at u a relative minimum with respect to the set U and if the constraints are qualified,
then there exist numbe rs λ
i
(u), 1 ≤i ≤m, such that the Kuhn-Tucker conditions
∇J(u)+
m

i=1
λ
i
(u)∇ϕ
i
(u) =0,
λ
i
(u) ≥0, 1 ≤i ≤m,
m

i=1
λ
i
(u)ϕ
i
(u) =0
(2.50)
are satisfied.

The convex constraints ϕ
i
in the above necessary condition are said to be qualified if
either all the functions ϕ
i
are affine and the set U is nonempt y, or there exists a point
w
∈ Ω such that for each i, ϕ
i
(w) ≤0 with strict inequality holding if ϕ
i
is not affine.
The solution to Problem 2.6 is given in the following theorem.
Theorem 2.9. One has
min
x∈Ᏹ
F(x) =0 =F(1,1,1, ,1,t) (2.51)
for any t
∈ [0,1].
Oscar Rojo 11
Proof. We observe that Ᏹ is a compact set and F isacontinuousfunctiononᏱ.Then,
there exists x
0
∈ Ᏹ such that F(x
0
) = min
x∈Ᏹ
F(x). The proof is based on the applica-
tion of the necessary condition given in the preceding theorem. In Problem 2.6,wehave
Ω

= V =
R
n
with the inner product x,y=

n
k
=1
x
k
y
k
, ϕ
i
(x) = h
i
(x), 1 ≤i ≤ n +1,U =
Ᏹ and J = F. T he functions h
i
,2≤ i ≤ n + 1, are linear. Therefore, they are convex and
affine. In addition, the function h
1
(x) = x
1
−1isaffine and convex and Ᏹ is nonempty.
Consequently, the functions h
i
,1≤i ≤ n + 1, are qualified. Moreover, these functions and
the objective function F are differentiable at any point in Ᏹ
−{e}. The gradients of the

constraint functions are
∇h
1
(x) =(1,0,0,0, ,0) = e
1
,
∇h
2
(x) =(−1,1,0,0, ,0),
∇h
3
(x) =(0,−1,1,0, ,0),
.
.
.
∇h
n−1
(x) =(0,0, ,0,−1,1,0),
∇h
n
(x) =(0,0, ,0,−1,1),
∇h
n+1
(x) =(0,0, ,0,−1).
(2.52)
Suppose that F has a relative minimum at x
∈ Ᏹ−{e} with respect to the set Ᏹ.Then,
there exist λ
i
(x) ≥ 0(forbrevityλ

i
= λ
i
(x)), 1 ≤ i ≤ n + 1, such that the Kuhn-Tucker
conditions
∇F(x)+
n+1

i=1
λ
i
∇h
i
(x) =0,
n+1

i=1
λ
i
h
i
(x) =0
(2.53)
hold. Hence,
∇F(x)+

λ
1
−λ
2

,λ
2
−λ
3
,λ
3
−λ
4
, , λ
n
−λ
n+1

=
0, (2.54)
λ
2

x
2
−1

+ λ
3

x
3
−x
2


+ ···+ λ
n

x
n
−x
n−1

+ λ
n+1

−
x
n

=
0. (2.55)
From (2.55), as λ
i
≥ 0, 1 ≤ i ≤ n +1,and0≤x
n
≤ x
n−1
≤···≤x
2
≤ 1, we have
λ
k

x

k−1
−x
k

=
0, 2 ≤k ≤ n, λ
n+1
x
n
= 0. (2.56)
Now, from (2.54),
n

k=1
∂
k
F(x)+λ
1
−λ
n+1
= 0. (2.57)
We will conclude that λ
1
= 0 by showing that the cases λ
1
> 0, x
n
> 0andλ
1
> 0, x

n
= 0
yield contradictions.
12 Inequalities on the mean and standard deviation
Suppose λ
1
> 0andx
n
> 0. In this case, λ
n+1
x
n
= 0 implies λ
n+1
= 0. Thus, (2.57)
becomes
n

k=1
∂
k
F(x) =−λ
1
< 0. (2.58)
We apply Lemma 2.7 to conclude that x is not one of the convex combinations in (2.26).
From (2.4),
x
=

1 −x

2

v
1
+

x
2
−x
3

v
2
+

x
3
−x
4

v
3
+ ···+

x
n−2
−x
n−1

v

n−2
+

x
n−1
−x
n

v
n−1
+ x
n
v
n
.
(2.59)
Then, there are at least two indexes i, j such that
1
=···=x
i
>x
i+1
=···=x
j
>x
j+1
. (2.60)
Therefore,
∂
1

F(x) =···=∂
i
F(x),
∂
i+1
F(x) =···=∂
j
F(x).
(2.61)
From (2.56), we get λ
i+1
= 0andλ
j+1
= 0. Now, from (2.54),
∂
i
F(x) =−λ
i
≤ 0,
∂
i+1
F(x) =λ
i+2
≥ 0,
∂
j
F(x) =−λ
j
≤ 0,
∂

n
F(x) =−λ
n
≤ 0.
(2.62)
The above equalities and inequalities together with (2.8)and(2.41)give
1
n

1 − f (x)

+
1
n(n −1)

1 −m

x
2

f

x
2

−
m

x
2


−
1 −m(x)
f (x) −m(x)

≤
0, (2.63)
1
n

1 − f (x)

+
1
n(n −1)

x
2
j
−m

x
2

f

x
2

−

m

x
2

−
x
j
−m(x)
f (x) −m(x)

=
0, (2.64)
1
n

1 − f (x)

+
1
n(n −1)

x
2
n
−m

x
2


f

x
2

−
m

x
2

−
x
n
−m(x)
f (x) −m(x)

≤
0. (2.65)
Subtracting (2.64)from(2.63)and(2.65), we obtain
1
−x
2
j
f

x
2

−

m

x
2

≤
1 −x
j
f

x
2

−
m

x
2

,
x
2
n
−x
2
j
f

x
2


−
m

x
2

≤
x
n
−x
j
f

x
2

−
m

x
2

.
(2.66)
Oscar Rojo 13
Dividing these inequalities by (1
−x
j
)and(x

n
−x
j
), respectively, we get
1+x
j
f

x
2

−
m

x
2

≤
1
f

x
2

−
m

x
2


,
x
n
+ x
j
f

x
2

−
a

x
2

≥
1
f

x
2

−
a

x
2

.

(2.67)
The last two inequalities imply x
n
≥ x
j
, which is contradiction.
Suppose now that λ
1
> 0andx
n
= 0. Let l be the largest index such t hat x
l
> 0. Thus,
x
l+1
= 0. From (2.55),
λ
2

x
2
−1

+ λ
3

x
3
−x
2


+ ···+ λ
l

x
l
−x
l−1

+ λ
l+1

−
x
l

=
0. (2.68)
Then,
λ
k

x
k−1
−x
k

=
0, 2 ≤k ≤ l, λ
l+1

x
l
= 0. (2.69)
Hence, λ
l+1
= 0. If l = n −1, then λ
n
= 0and∂
n
F(x) =λ
n+1
≥ 0. If l ≤ n −2, then ∂
l
F(x) =
−
λ
l
≤ 0. In both situations, we conclude that x is not one of the convex combinations in
(2.26). Therefore, there are at least two indexes i, j such that
1
=···=x
i
>x
i+1
=···=x
j
>x
j+1
. (2.70)
Now, we repeat the argument used above to get that x

l
≥ x
j
, which is a contradiction.
Consequently, λ
1
= 0. From (2.57),
n

k=1
∂
k
F(x) =λ
n+1
≥ 0. (2.71)
We apply now Lemma 2.7 to conclude that x is one of the convex combinations in (2.26).
Let x
= x
N
(t) =te+(1 −t)v
N
,1≤ N ≤ n −2, and t ∈ [0,1). Then, x
1
= x
2
=···=x
N
= 1,
x
N+1

= x
N+2
=···=x
n
= t,andh
N+1
(x) = t −1 < 0. From (2.56), we obtain λ
N+1
= 0.
Thus, from (2.54), ∂
N+1
F(x) =λ
N+2
≥ 0. This contradicts (2.40). Thus, x = x
N
(t)forN =
1,2, ,n−2andt ∈ [0,1). Consequently, x =x
n−1
(t) =(1,1, ,1,t)forsomet ∈ [0,1).
Finally,
F(1,1, ,1,t)
= f

1,1, ,1,t
2

−

f (1,1, ,1,t)


2
= 1 −1 =0 (2.72)
for any t
∈ [0,1]. Hence, min
x∈Ᏹ
F(x) = 0 = F(1,1, ,1,t)foranyt ∈ [0,1]. Thus, the
theorem has been proved.

Theorem 2.10. If y
1
≥ y
2
≥ y
3
≥···≥y
n
≥ 0, then
m

y
2
p

+
1
√
n −1
s

y

2
p

≤

m

y
2
p+1

+
1
√
n −1
s

y
2
p+1

, (2.73)
14 Inequalities on the mean and standard deviation
that is,

n
k
=1
y
2

p
k
n
+
1

n(n −1)






n

k=1
y
2
p+1
k
−


n
k
=1
y
2
p
k


2
n
≤
⎡
⎢
⎢
⎢
⎣

n
k
=1
y
2
p+1
k
n
+
1

n(n −1)






n


k=1
y
2
p+2
k
−


n
k
=1
y
2
p+1
k

2
n
⎤
⎥
⎥
⎥
⎦
1/2
(2.74)
for p
= 0,1,2, The equality holds if and only if y
1
= y
2

=···=y
n−1
.
Proof. If y
1
= 0, then y
2
= y
3
=···= y
n
= 0 and the theorem is immediate. Hence, we
assume that y
1
> 0. Let p be a nonnegative integer and let x
k
= y
k
/y
1
for k = 1,2, ,n.
Clearly, 1
= x
2
p
1
≥ x
2
p
2

≥ x
2
p
3
≥···≥x
2
p
n
≥ 0. From Theorem 2.9,wehave

f

1,x
2
p
2
,x
2
p
3
, , x
2
p
m

2
≤ f

1,x
2

p+1
2
,x
2
p+1
3
, , x
2
p+1
m

, (2.75)
that is,
⎛
⎜
⎜
⎜
⎝
1+

n
k
=2
x
2
p
k
n
+
1


n(n −1)






1+
n

k=2
x
2
p+1
k
−

1+

n
j
=2
x
2
p
j

2
n

⎞
⎟
⎟
⎟
⎠
2
≤
1+

n
k
=2
x
2
p+1
k
n
+
1

n(n −1)






1+
n


k=2
x
2
p+2
k
−

1+

n
j
=2
x
2
p+1
j

2
n
(2.76)
with equality if and only if x
1
= x
2
=···=x
n−1
.Multiplyingbyy
2
p+1
1

, the inequality in
(2.74) i s obtained with equality if and only if y
1
= y
2
=···=y
n−1
. This completes the
proof.

Corollar y 2.11. Let y
1
≥ y
2
≥ y
3
≥···≥y
n
≥ 0. Then (l
2
p
(y))
∞
p=0
,
l
2
p
(y) =
⎛

⎝

y 
2
p
2
p
n
+
1

n(n −1)



 y
2
p+1
2
p+1
−

y 
2
p+1
2
p
n
⎞
⎠

2
−p
=

m

y
2
p

+
1
√
n −1
s

y
2
p


2
−p
,
(2.77)
is an strictly increasing sequence converging to y
1
except if y
1
= y

2
=···=y
n−1
. In this case,
l
2
p
(y) = y
1
for all p.
Oscar Rojo 15
Proof. We know that (l
2
p
(y))
∞
p=0
is a sequence of lower bounds for y
1
.FromTheorem 2.1,
this sequence converges to y
1
. Applying inequality (2.74), we obtain
⎛
⎜
⎜
⎜
⎝

n

k
=1
y
2
p
k
n
+
1

n(n −1)






n

k=1
y
2
p+1
k
−


n
j
=1

y
2
p
j

2
n
⎞
⎟
⎟
⎟
⎠
2
≤

n
k
=1
y
2
p+1
k
n
+
1

n(n −1)







n

k=1
y
2
p+2
k
−


n
j
=1
y
2
p+1
j

2
n
.
(2.78)
Therefore, l
2
p+1
2
p

(y) ≤l
2
p+1
2
p+1
(y), that is, l
2
p
(y) ≤l
2
p+1
(y). The equality in all the above inequal-
ities takes place if and only if λ
1
=y
2
=···=y
n−1
. In this case, l
2
p
(y)=λ
1
for all p. 
Acknowledgment
This work is supported by Fondecyt 1040218, Chile.
References
[1] P.G.Ciarlet,Introduction to Numerical Linear Algebra and Optimisation, Cambridge Texts in
Applied Mathematics, Cambridge University Press, Cambridge, 1991.
[2] O. Rojo and H. Rojo, A decreasing sequence of upper bounds on the largest Laplacian eigenvalue of

a graph, Linear Algebra and Its Applications 381 (2004), 97–116.
[3] H. Wolkowicz and G. P. H. Styan, Bounds for eigenvalues using traces, Linear Algebra and Its
Applications 29 (1980), 471–506.
Oscar Rojo: Departamento de Matem
´
aticas, Universidad Cat
´
olica del Norte, Casilla 1280,
Antofagasta, Chile
E-mail address: