Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 3, Issue 3, Article 43, 2002
INEQUALITIES ON LINEAR FUNCTIONS AND CIRCULAR POWERS
PANTELIMON STĂNICĂ
AUBURN U NIVERSITY M ONTGOMERY,
D EPARTMENT OF M ATHEMATICS ,
M ONTGOMERY, AL 36124-4023, USA.
stanica@strudel.aum.edu
URL: http://sciences.aum.edu/˜ stanpan
Received 25 February, 2002; accepted 10 April, 2002
Communicated by C.P. Niculescu
A BSTRACT. We prove some inequalities such as
x
x
F (x1 σ(1) , . . . , xnσ(n) ) ≤ F (xx1 1 , . . . , xxnn ),
where F is a linear function or a linear function in logarithms and σ is a permutation, which
is a product of disjoint translation cycles. Stronger inequalities are proved for second-order
recurrence sequences, generalizing those of Diaz.
Key words and phrases: Binary Sequences, Fibonacci Numbers, Linear Forms, Inequalities.
2000 Mathematics Subject Classification. 11B37, 11B39, 26D15.
1. I NTRODUCTION
Define the second-order recurrent sequence by
(1.1)
xn+1 = axn + bxn−1 , x0 ≥ 0, x1 ≥ 1,
with a, b ≥ 1. If a = b = 1 and x0 = 0, x1 = 1 (or x0 = 2, x1 = 1), then xn is the Fibonacci
sequence, Fn (or Pell sequence, Pn ). Inequalities on Fibonacci numbers were used recently
by Bar-Noy et.al [1], to study a 9/8− approximation for a variant of the problem that models
the Broadcast Disks application (model for efficient caching of web pages). In [2], J.L. Diaz
proposed the following two inequalities:
Fn+2
Fn+2
Fn+1
Fn
(a) FnFn+1 + Fn+1
+ Fn+2
< FnFn + Fn+1
+ Fn+2
,
Fn+1 Fn+2 Fn
Fn Fn+1 Fn+2
(b) Fn Fn+1 Fn+2 < Fn Fn+1 Fn+2 .
In this note we show that the inequalities proposed by Diaz are not specific to the Fibonacci
sequence, holding for any strictly increasing sequence. Moreover, we prove that stronger inequalities hold for any second-order recurrent sequence as in (1.1). Furthermore, we pose a
problem for future research.
ISSN (electronic): 1443-5756
c 2002 Victoria University. All rights reserved.
The author would like to thank Professor C.P. Niculescu for his helpful comments, which improved significantly the presentation.
015-02
2
PANTELIMON S TĂNIC Ă
2. T HE R ESULTS
We wondered if the inequalities (a), (b) were dependent on the Fibonacci sequence or if they
can be extended to binary recurrent sequences. From here on, we assume that all sequences
have positive terms. Without too great a difficulty, we prove, for a binary sequence, that
Theorem 2.1. For any positive integer n,
x
x
x
n+2
n+1
n+2
n
,
+ xn+2
+ xxn+2
< xxnn + xn+1
xxnn+1 + xn+1
x
x
x
n+2
n+1
n+2 xn
.
xn+2
xn+2 < xxnn xn+1
xxnn+1 xn+1
(2.1)
(2.2)
Proof. We shall prove
(2.3)
xy + y ax+by + (ax + by)x < xx + y y + (ax + by)ax+by ,
if 0 < x < y, which will imply our theorem. For easy writing, we denote z = ax + by. Then
(2.3) is equivalent to
(2.4)
xx xy−x − 1 + y y y z−y − 1 < z y z z−y − z −(y−x) .
Now, xx + y y < xy + y y < (x + y)y ≤ z y , since a, b ≥ 1. Moreover,
xy−x − 1 + y z−y − 1 = xy−x + y z−y − 2
< xz−y + y z−y − 1
< (x + y)z−y − 1
< z z−y − z −(y−x) .
Taking A = xx , B = y y , C = xy−x − 1, D = y z−y − 1 and using the inequality for positive
numbers AC + BD ≤ (A + B)(C + D), we obtain (2.4).
The inequality (2.2) is implied by
xy y z z x < xx y y z z ⇐⇒ xy−x y z−y < z z−x .
(2.5)
But z z−x = z (z−y)+(y−x) ≥ (x + y)z−y (x + y)(y−x) > y z−y xy−x . The theorem is proved.
Remark 2.2. We preferred to give this proof since it can be seen that the two inequalities are
far from being tight. We remark that inequality (2.2) can be also shown by using Theorem 2.7.
With a little effort, while not attempting to have the best bound, we can improve it, and also
prove that the gaps are approaching infinity.
Theorem 2.3. We have
xn+1
xn+2
xn+2
n
+ xn+2
− xn+2 xn+1 −xn
< xxnn + xn+1
+ xxn+2
xxnn+1 + xn+1
x
x
x
x
n+1 xn
n+1
n+2
n+2 xn
xn+2 .
xn+2
− 3xxnn xn+1
xn+2 < xxnn xn+1
xxnn+1 xn+1
In particular,
xn+2
xn+1
xn+2 n
]=∞
+ xxn+2
+ xn+2
− xxnn+1 + xn+1
lim [ xxnn + xn+1
n→∞
x
x
x
n+2 xn
n+1
n+2
xn+2 ] = ∞.
xn+2
− xxnn+1 xn+1
lim [xxnn xn+1
n→∞
In fact, the inequalities (2.1), (2.2) are not dependent on binary sequences, at all. A much
more general statement is true. Take σ a permutation, which is a product of disjoint cyclic
(translations by a fixed number, c(i) = i + t) permutations.
Theorem 2.4. Let n ≥ 2 and 1 ≤ x1 < x2 < . . . < xn a strictly increasing sequence. Then
n
n
X
X
xσ(i)
(2.6)
xi
≤
xxi i ,
i=1
J. Inequal. Pure and Appl. Math., 3(3) Art. 43, 2002
i=1
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I NEQUALITIES ON L INEAR F UNCTIONS AND C IRCULAR P OWERS
3
and
n
Y
(2.7)
x
xi σ(i)
≤
i=1
n
Y
xxi i ,
i=1
with strict inequality if σ is not the identity.
Proof. If σ is the identity permutation, the equality is obvious. Now, assume that σ(i) = i + t.
We take the case of t = 1 (the others are similar). We prove (2.6) by induction on n. If n = 2,
we need
xx1 2 + xx2 1 < xx1 1 + xx2 2 ,
which is equivalent to
xx1 1 xx1 2 −x−1 − 1 < xx2 1 xx2 2 −x−1 − 1 .
The last inequality is certainly valid, since xx1 1 < xx2 1 and xx1 2 −x1 − 1 < xx2 2 −x1 − 1.
Assuming the statement holds true for n, we prove it for n + 1. We need
n+1
X
(2.8)
x
xi i+1
<
i=1
n+1
X
xxi i ,
i=1
where xn+2 := x1 . We re-write (2.8) as
x
n+1
1
(xx1 2 + xx2 3 + · · · + xxn1 ) + xnxn+1 + xxn+1
− xxn1 < xx1 1 + xx2 2 + · · · + xxnn + xn+1
,
and using induction, it suffices to prove that
x
n+1
1
xnxn+1 + xxn+1
− xxn1 < xn+1
.
The previous inequality is equivalent to
xn+1 −x1
1
xxn1 xxnn+1 −x1 − 1 < xxn+1
xn+1
−1 ,
which is obviously true, since xn < xn+1 .
The inequality (2.7) (when σ(i) = i + t) can be proved by induction, as well. If n = 2, then
xx1 2 xx2 1 < xx1 1 xx2 2 ⇐⇒ x1x2 −x1 < x2x2 −x1 ,
which is true since x1 < x2 . Assuming the inequality holds true for n, we prove it for n + 1.
We need
xn+1
n
1
xx1 2 · · · xxnn+1 xxn1 = xx1 2 · · · xxn−1
xxn1 xxnn+1 −x1 xxn+1
< xx1 1 · · · xn+1
.
Using the induction step, it suffices to prove
x
x
n+1
n+1
1
< xn+1
⇐⇒ xxnn+1 −x1 < xn+1
xxnn+1 −x1 xxn+1
−x1
,
which is valid since xn < xn+1 .
Now, take the general permutation σ 6=identity, which is a product of disjoint cyclic permutations. Thus, σ can be written as a product of disjoint cycles as σ = C1 × C2 × · · · × Cm .
Recall that σ was taken such that all of its cycles Ck are translations by a fixed number, say tk .
Take Ck a cycle of length ek and choose an index in Ck , say ik . Since σ is not the identity, then
there is an index k such that ek 6= 1. We write the inequalities (2.6) and (2.7) as
m eX
k −1
X
xσj+1 (i
k)
xσj (ik )
<
k=1 j=0
m eY
k −1
Y
k=1 j=0
J. Inequal. Pure and Appl. Math., 3(3) Art. 43, 2002
m eX
k −1
X
xσj (i
)
xσj (ikk) ,
k=1 j=0
xσj+1 (i
xσj (ik )
k)
<
m eY
k −1
Y
xσj (i
)
xσj (ikk) ,
k=1 j=0
http://jipam.vu.edu.au/
4
PANTELIMON S TĂNIC Ă
Therefore, it suffices to prove that, for any k, with ek 6= 1, we have
eX
k −1
xσj+1 (i
xσj (ik )
k)
<
eX
k −1
j=0
j=0
eY
k −1
eY
k −1
xσj+1 (i
xσj (ik )
k)
<
j=0
xσj (i
)
xσj (i
)
xσj (ikk) ,
xσj (ikk) ,
j=0
that is,
eX
k −1
xσj (i
xσj (ikk)
)+tk
<
eX
k −1
j=0
j=0
eY
k −1
eY
k −1
xσj (i
xσj (ikk)
)+tk
<
j=0
xσj (i
)
xσj (i
)
xσj (ikk) ,
xσj (ikk) .
j=0
For k fixed, the above inequalities are just applications of the previous step (of σ(i) = i + t), by
taking a sequence yl to be xσj (ik ) in increasing order (we could take from the beginning ik to be
the minimum index in each cycle Ck ).
We can slightly extend the previous result (for a similar permutation σ) in the following (we
omit the proof).
Theorem 2.5. For any increasing sequence 0 < x1 < · · · < xn , we have
n
X
(2.9)
x
n
X
i=1
i=1
n
Y
x
n
Y
ai xi σ(i) ≤
ai xi σ(i) ≤
ai xxi i , and
ai xxi i ,
i=1
i=1
where ai ≥ 0.
A parallel result involving logarithms is also true (σ is a permutation as before).
Theorem 2.6. For any finite increasing sequence, 0 < x1 < x2 < · · · < xn , and any positive
real numbers ai , we have
n
X
i=1
n
Y
ai xσ(i) log(xi ) ≤
ai xσ(i) log(xi ) =
i=1
n
X
ai xi log(xi ), and
i=1
n
Y
ai xi log(xi ).
i=1
The second identity is easily true since every ai , xi and log xi occurs in both sides. We omit
the proof of the first inequality, since it can be deduced easily (as the referee observed) from the
known fact (see [3, p. 261])
Theorem 2.7. Given two increasing sequences u1 ≤ u2 ≤ · · · ≤ un and w1 ≤ w2 ≤ · · · ≤ wn ,
then
n
n
n
X
X
X
ui wn+1−i ≤
uτ (i) wσ(i) ≤
ui wi ,
i=1
i=1
i=1
for any permutations σ, τ .
J. Inequal. Pure and Appl. Math., 3(3) Art. 43, 2002
http://jipam.vu.edu.au/
I NEQUALITIES ON L INEAR F UNCTIONS AND C IRCULAR P OWERS
5
3. F URTHER C OMMENTS
We believe that other inequalities of the type occurring in our theorems can also be constructed. Let F : Rn → R be a function, with the properties
(3.1)
If xi ≤ yi , i = 1, . . . , n, then F (x1 , . . . , xn ) ≤ F (y1 , . . . , yn ),
with strict inequality if there is an index i such that xi < yi .
and
For 0 < x1 < x2 < · · · < xn , then, F (xx1 2 , xx2 3 , . . . , xxn1 ) ≤ F (xx1 1 , xx2 2 , . . . , xxnn ).
Pn
As examples, we have the linear
i=1 ai xi , the linear form in
Pn polynomial F (x1 , . . . , xn ) =
logarithms F (x1 , . . . , xn ) = i=1 ai log(xi ), and the corresponding products.
We ask for more examples of functions satisfying (3.1) and (3.2), which cannot be derived
trivially from the previous examples (by raising each variable to the same power, for instance).
Is it true that any symmetric polynomial satisfies (3.1) and (3.2)? In addition to more examples, it might be worth investigating the general form of polynomial functions that satisfy these
properties.
This looks like a mathematical version of the philosophy saying:
Going one step at the time it is far better than jumping too fast and then at the end falling to the
bottom.
(3.2)
R EFERENCES
[1] A. BAR-NOY, R. BHATIA, J. NAOR AND B. SCHIEBER, Minimizing Service and Operating Costs
of Periodic Scheduling, Symposium on Discrete Algorithms (SODA), p. 36, 1998.
[2] J.L. DIAZ, Problem H-581, Fibonacci Quart., 40(1) (2002).
[3] G. HARDY, J.E. LITTLEWOOD, G. PÒLYA, Inequalities, Cambridge Univ. Press, 2001.
J. Inequal. Pure and Appl. Math., 3(3) Art. 43, 2002
http://jipam.vu.edu.au/