Proposed Proof of the Riemann Hypothesis

. For every prime number q n , we define the inequality (cid:81) qq , where θ ( x ) is the Chebyshev function and γ ≈ 0 . 57721 is the Euler-Mascheroni constant. This is known as the Nicolas inequality. The Nicolas criterion states that the Riemann hypothesis is true if and only if the Nicolas inequality is satisfied for all primes q n > 2 . We prove indeed that the Nicolas inequality is satisfied for all primes q n > 2 . In this way, we show that the Riemann hypothesis is true.


INTRODUCTION
The Riemann hypothesis is a conjecture that the Riemann zeta function has its zeros only at the negative even integers and complex numbers with real part 1 2 [1]. In mathematics, the Chebyshev function θ(x) is given by where q ≤ x means all the prime numbers q that are less than or equal to x. We define a sequence based on this function. Definition. For every prime number q n , we define the sequence of real numbers: Say Nicolas(q n ) holds provided The constant γ ≈ 0.57721 is the Euler-Mascheroni constant and log is the natural logarithm. The importance of this inequality is: Theorem 1. [4]. Nicolas(q n ) holds for all prime numbers q n > 2 if and only if the Riemann hypothesis is true.
Besides, we define the following value, Putting all together yields the proof that the Nicolas inequality is satisfied for all prime numbers greater than 2. Consequently, we prove that the Riemann hypothesis is true.

A CENTRAL LEMMA
The following is a key lemma. Lemma 1. There exists a natural number N such that X n < e γ ×6 π 2 + ε for all natural numbers n > N and ε ≤ 6 π 2 . Proof. The limit superior of a sequence of real numbers y n is the smallest real number b such that, for any positive real number ε, there exists a natural number N such that y n < b + ε for all natural numbers n > N . Therefore, this is a consequence of the theorem 2.

PROOF OF MAIN THEOREM
Theorem 4. The Riemann hypothesis is true.
Proof. From the lemma 1, we know that there exists a natural number N such that X n < e γ ×6 π 2 + ε for all natural numbers n > N and ε ≤ 6 π 2 . We multiply the both sides of the inequality for the constant c = ε × π 2 6 due to From the theorem 3, we note that q≤qn q 2 q 2 −1 × 6 π 2 is strictly increasing as q n increases. Besides, we have that Proposition. We state the following proposition S: There exists a natural number m such that for a sufficiently small constant 0 < c ≤ 1. Hence, we could have for some small constant 0 < d ≤ 1. This implies that Hence, Nicolas(q m ) would not hold.
Proposition. We state another proposition T : The Riemann hypothesis is false.
So, we would have the implication S ⇒ T should be true because of the theorem 1. However, we know that and thus, we would get e γ e γ + c .
Following the previous steps, we would obtain that Nicolas (3) does not hold. In this way, we obtain a contradiction since Nicolas(3) holds indeed. Consequently, the implication S ⇒ T cannot be true. If the implication S ⇒ T is false, then T is also false. So, the proposition T which exactly states that: The Riemann hypothesis is false cannot be true. By contraposition, we can conclude that the Riemann hypothesis is indeed true.