WHEN THE ROBIN INEQUALITY DOES NOT HOLD

. In mathematics, the Riemann Hypothesis is a conjec-ture that the Riemann zeta function has its zeros only at the nega-tive even integers and complex numbers with real part 12 . It is one of the seven Millennium Prize Problems selected by the Clay Mathematics Institute to carry a US 1,000,000 prize for the ﬁrst correct solution. In 1915, Ramanujan proved that under the assumption of the Riemann Hypothesis, the inequality σ ( n ) < e γ × n × log log n holds for all suﬃciently large n , where σ ( n ) is the sum-of-divisors function and γ ≈ 0 . 57721 is the Euler-Mascheroni constant. In 1984, Guy Robin proved that the inequality is true for all n > 5040 if and only if the Riemann Hypothesis is true. Let n > 5040 be n = r × q , where q denotes the largest prime factor of n . If n > 5040 is the smallest number such that Robin inequality does not hold, then we show the following inequality is also satisﬁed: q √ e + loglog r loglog n > 2.


Introduction
As usual σ(n) is the sum-of-divisors function of n [Cho+07]: d|n d.
Define f (n) to be σ(n) n . Say Robins(n) holds provided f (n) < e γ × log log n.
The constant γ is the Euler-Mascheroni constant, and log is the natural logarithm. The importance of this property is: If Robins(n) holds for all n > 5040, then the Riemann Hypothesis is true [Rob84].
There are several known results about the possible counterexamples of Robins(n) when n > 5040 [Cho+07]. In addition, we show that Theorem 1.2.
[counterexample] Let n > 5040 be n = r × q, where q denotes the largest prime factor of n. If n > 5040 is the smallest number such that Robins(n) does not hold, then q √ e + log log r log log n > 2.

Some Useful Lemmas
The following lemma is a very helpful inequality: We have Proof. We know 1 + x ≤ e x [Koz21]. Therefore, However, for every real number y ∈ R [Koz21]: y × e y ≥ y + y 2 + y 3 2 and this can be transformed into 1 y × e y ≤ 1 y + y 2 + y 3 2 .
Consequently, we show Here, it is another practical result: Suppose that n > 5040 and let n = r × q, where q denotes the largest prime factor of n. We have Proof. Suppose that n is the form of m × q k where q m and m and k are natural numbers. We have that since f is multiplicative and m and q are coprimes [Voj20]. However, we know that In this way, we obtain that according to the value of f (q) [Voj20]. In addition, we analyze that because f is multiplicative and m and q are coprimes [Voj20]. Finally, we obtain that and as a consequence, the proof is completed.

Proof of Main Theorem
Theorem 3.1. Let n > 5040 be n = r × q, where q denotes the largest prime factor of n. If n > 5040 is the smallest number such that Robins(n) does not hold, then q √ e + log log r log log n > 2.
Proof. Suppose that n is the smallest integer exceeding 5040 that does not satisfy the Robin's inequality. Let n = r × q, where q denotes the largest prime factor of n. In this way, the following inequality f (n) ≥ e γ × log log n should be true. We know that (1 + 1 q ) × e γ × log log r > e γ × log log n should be true as well. Certainly, if n is the smallest counterexample exceeding 5040 of the Robin's inequality, then Robins(r) holds [Cho+07]. That is the same as (1 + 1 q ) × log log r > log log n.
We have that (1 + 1 q ) × log log r > log(log r + log q) where we notice that log(a + c) = log a + log(1 + c a ). This follows (1 + 1 q ) × log log r > log log r + log(1 + log q log r ) which is equal to (1 + q) × log log r > q × log log r + q × log(1 + log q log r ) and thus, log log r > q × log(1 + log q log r ).
This implies that log log r log(1 + log q log r ) = log log r log log r+log q log r = log log r log log n log r = log log r log log n − log log r = log log r log log n × (1 − log log r log log n ) = log log r log log n (1 − log log r log log n ) > q should be true. If we assume that y = 1 − log log r log log n , then we analyze that 1 y + y 2 + y 3 2 ≥ log log r log log n (1 − log log r log log n ) because of lemma 2.1 [ineq]. As result, we have that 1 y + y 2 + y 3 2 > q and therefore, 1 1 + y + y 2 2 > q × y.
Consequently, we obtain that 1 > q × y which is the same as e > e q×y . Because of we have that 1 + y ≤ e y [Koz21], then e > e q×y ≥ (1 + y) q = (2 − log log r log log n ) q that is q √ e > (2 − log log r log log n ) and finally, q √ e + log log r log log n > 2.