On certain bounds for the divisor function

: We offer various bounds for the divisor function d ( n ) , in terms of n, or other arithmetical functions.


Introduction
For a positive integer n > 1, let d(n) denote the number of all positive divisors of n.Our aim in what follows will be to offer bounds for d(n) in terms of functions of n, or other arithmetic functions, such as the Euler totient function φ(n) or σ(n), the sum of divisors of n.We will use also the function ω(n) denoting number of distinct prime factors of n.
If n = p a 1 1 • • • p ar r (p i distinct primes, a i positive integers) is the prime factorization of n, then it is well-known that where a i ≥ 1 (i = 1, 2, . . ., r); so (1) immediately gives A classical upper bound for d(n) is while it is also known that (see [6]) If n is composite, then it is well-known that [8] where φ(n) denotes the Euler totient function.Thus, from ( 4) and ( 5) we get composite, so we get In [7] it has been shown that and this clearly improves (3) of ( 4) for sufficiently large values of n.
In what follows, we will offer other bounds which are more precise for certain values of n.

Main results
Theorem 1.One has for n ≥ 2, where the second inequality holds true for n ̸ = 6.
Proof.As ω(n) ≥ 1, the first inequality of (8) follows by σ(n This last inequality follows, e.g., by σ(n) ≥ ψ(n), where denotes the Dedekind arithmetical function.By the algebraic inequality r i=1 (1 for x i ∈ (0, 1) the result immediately follows (by letting For the second inequality of (8), use the following result of the author [4]: for any n ≥ 2, n ̸ = 6; with equality only for n = 10 or n = prime.
Finally, the third inequality of ( 8) is a consequence of another result by the author [3]: This follows by n + 1 − 2 √ n = ( √ n − 1) 2 , and simple computations.The last inequality of ( 8) is trivial, but we note that it improves the classical result d(n) < due to Sivaramakrishnan and Venkataraman.
Remark 1. Particularly, we get the new inequality or written equivalently : for n ̸ = 6; n ≥ 2.
Theorem 2. One has for any n ≥ 2; where σ 2 (n) denotes the sum of squares of divisors of n.
Proof.Let a i (i = 1, n) be positive real numbers, and let A n , respectively, G n denote their arithmetic, respectively, geometric means of (a i ).
In paper [5] is proved among others the inequalities: where k ≥ 2; and . After some elementary computations, from (14), we get inequality (13), where the inequality (13) is considered for k (in place of n.) Remark 2. From the first inequality of (13) we get We now consider extensions of inequalities of type (7).We will use the method of [7].One has the following similar inequalities.
Theorem 3. One has for any n ≥ 2.
The following auxiliary result will be used: where for all a ≥ 1, p ≥ 2; k ≥ 2.
Proof of Lemma 1.Let us consider the real variable function f (a) = (a + 1) • p −a/k .This function has a derivative If is immediate that a 0 is maximum point to f, the maximum being attained at a 0 .We get that f (a) ≤ f (a 0 ) = M k (p), so inequality (21) follows.
As a + 1 = d(p a ), we get from (21) that This completes the proof of the Lemma.□ Proof of Theorem 3. Now, remark that p a > (a + 1) 4 follows by p a ≥ 17 a if 17 a > (a + 1) 4 .Since 17 1 > 2 4 = 16, by an easy induction argument the inequality holds true for any a ≥ 1.As For the proof of (17) we will use the inequality 37 a/5 > a + 1, and proceed in the same manner, as in the case k = 4.
A computer computation gives us 2≤p≤31 M 5 (p) ≈ 33.4725 . . ., and the inequality (17) follows.The proofs of ( 18) and ( 19) can be derived in the same manner, and we omit the details.□ Remark 3. If n is odd, the above inequalities can be further sharpened.For example, when k = 3, (when M 3 (2) = 1) we get and for k = 4 and 5, Remark 4. The method of Theorem 3 can be used also for k = 2, and as which further improve relation (3).
Proof.We will use the classical Chebysheff inequality for the sequences (x i ) and (y i ) having the property (x i − x j ) • (y i − y j ) ≥ 0 for all i, j ∈ {1, 2, . . ., n}.Let can be applied.After simple computations, we get relation (28).
Remark 5.For k = l we get from (28) where the last inequality is well-known.
Lemma 3. Let (a i ), (b i ), (i = 1, 2, . . ., n) be two sequences with the property (a i −a j )(b i −b j ) ≤ 0 (i, j ∈ {1, 2, . . ., n}).Then one has the inequalities and Proof.First remark that, the converse of inequality ( 29) is true now, so the first inequality of (31); and also the second inequality of (32) (applied to y 2 i applied first to x i = a i , then x i = b i gives the second inequality of (31), while applying it to x i = a i b i we get the first inequality of (32).
where σ a (m) denotes the sum of a-th powers of divisors of m.
where the first inequality of (38) is trivial.

Remark 6 .a i b i 2 ≤Theorem 4 .
Both inequalities offer a refinement of the classical Cauchy inequality n i=1 For any m ≥ 2 and k, l > 0 one has