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OCTOGON MATHEMATICAL MAGAZINE Vol. 17, No.1, April 2009, pp 288-290

ISSN 1222-5657, ISBN 978-973-88255-5-0, www.hetfalu.ro/octogon

288

Improvement of one of S´ andor‘s inequalities

Nicu¸sor Minculete32

ABSTRACT.The objective of this paper is to present an improvement of S´andor‘s inequality

σk(n)·σ1(n) σk1

2 (n) ≤n−(k4l) ·n

k+l2 +1

2 , for anyn, k, l∈N, whereσk(n)is the sum ofkth powers of divisors of n, so σk(n) =P

d|n

dk.

INTRODUCTION

Letn be a positive integer,n≥1. We note with σk(n) the sum ofkth powers of divisors ofn, so, σk(n) =P

d|n

dk, whence we obtain the following equalities: σ1(n) =σ(n) and σ0(n) =τ(n)−the number of divisors of n.

In [1], J. S´andor shows that pσk(n)·σl(n)

σkl

2 (n) ≤n−(k4l) ·nk+l2 + 1

2 , for anyn, k, l∈N (1) In [2] an inequality which is due to J.B. Diaz and F.T. Metcalf is proved, namely:

Lemma 1.1 Letn be a positive integer,n≥2. For every

a1, a2, ..., an∈R and for everyb1, b2, ..., bn∈R with m≤ abii ≤M and m, M ∈R, we have the following inequality:

Xn i=1

a2i +mM Xn i=1

b2i ≤(m+M) Xn

i=1

aibi. (2)

32Received: 25.03.2009

2000Mathematics Subject Classification. 11A25

Key words and phrases. The sum of the natural divisors of n, the sum of kth powers of divisors ofn

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Improvement of one of S´andor‘s inequalities 289 1. MAIN RESULT

Theorem 1.2. For everyn, k, l∈N withn≥2 and k2l ∈N the following relation

k(n)·σl(n) σkl

2 (n) ≤ nl4kσk(n) +nk4lσl(n) 2σkl

2 (n) ≤n(k4l) ·nk+l2 + 1

2 , is true. (3) Proof. In the Lemma 1.1, making the substitutionai =q

dki and bi = √1

dli, wheredi is the divisor of n, for anyi= 1, τ(n). Since

1≤ abii = q

dk+li ≤nk+l2 and aibi =d

kl

i2 , we take m= 1 and M =nk+l2 . Therefore, inequality (2) becomes

τ(n)X

i=1

dki +nk+l2

τ(n)X

i=1

1 dli

1 +nk+l2 τ(n)X

i=1

d

kl

i2

which is equivalent to

σk(n) +nk+l2 σl(n) nl

1 +nk+l2 σkl

2 (n) so that

σk(n) +nk2lσl(n)≤

1 +nk+l2 σkl

2 (n), (4)

for everyn, k, l∈N with n≥2.

The arithmetical mean is greater than the geometrical mean or they are equal, so for everyn, k, l∈N withn≥2, we have

q

nk2lσk(n)σl(n)≤ σk(n) +nk2lσl(n)

2 . (5)

Consequently, from the relations (4) and (5), we deduce the inequality pσk(n)σl(n)

σkl

2 (n) ≤ nl4kσk(n) +nk4lσl(n) 2σkl

2 (n) ≤n−(k4l) ·nk+l2 +l

2 .

Remark Fork→k+ 2 andl→k we obtain the relationk+2(n)·σk(n)

σ(n) ≤

1

nσk+2(n) +√

k(n)

2σ(n) ≤ 1

√n ·nk+1+ 1

2 , (6)

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290 Nicu¸sor Minculete for everyn, k∈N with n≥2.

Fork=l, we deduce another inequality which is due to S´andor, namely, σk(n)

τ(n) ≤ nk+ 1

2 , (7)

for everyn, k∈N withn≥2.

REFERENCES

[1] S´andor, J., On Jordan’s Arithmetical Function, Gazeta Matematic˘a nr.

2-3/1993.

[2] Drimbe, M.O.,Inegalit˘at¸i. Idei ¸si metode, Editura GIL, Zal˘au, 2003.

“Dimitrie Cantemir” University of Bra¸sov E-mail: minculeten@yahoo.com

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