On Proving Ramanujan’s Inequality using a Sharper Bound for the Prime Counting Function π(x)

1. Introduction

The notion of analyzing the proportion of prime numbers over the real line $\mathbb{R}$ first came into the limelight thanks to the genius work of one of the greatest and most gifted mathematicians of all time named Srinivasa Ramanujan, as evident from his letters ([12], pp. xxiii-xxx , 349-353) to another one of the most prominent mathematicians of 20^th century, G. H. Hardy during the months of Jan/Feb of 1913, which are testaments to several strong assertions about the Prime Counting Function, $\pi \left(x\right)$ (cf. Definition $2.1.1$ [15]).

In the following years, Hardy himself analyzed some of those results ([13,14], pp. 234-238), and even wholeheartedly acknowledged them in many of his publications, one such notable result is the Prime Number Theorem (cf. Theorem $2.1.1$ [15]).

Ramanujan provided several inequalities regarding the behavior and the asymptotic nature of $\pi \left(x\right)$. One of such relation can be found in the notebooks written by Ramanujan himself has the following claim.

Theorem 1. 

(Ramanujan’s Inequality [1])). For x sufficiently large, we shall have,

$$\begin{array}{c}\hfill {\left(\pi \left(x\right)\right)}^2<{\displaystyle \frac{ex}{logx}}\pi \left({\displaystyle \frac{x}{e}}\right)\end{array}$$

(1)

Worth mentioning that, Ramanujan indeed provided a simple, yet unique solution in support of his claim. Furthermore, it has been well established that, the result is not true for every positive real x. Thus, the most intriguing question that the statement of Theorem (1) poses is, is there any $x_0$ such that, Ramanujan’s Inequality will be unconditionally true for every $x\ge x_0$?

A brilliant effort put up by F. S. Wheeler, J. Keiper, and W. Galway in search for such $x_0$ using tools such as MATHEMATICA went in vain, although independently Galway successfully computed the largest prime counterexample below ${10}^{11}$ at $x=38358837677$. However, Hassani ([3], Theorem 1.2) proposed a more inspiring answer to the question in a way that, ∃ such $x_0=138766146692471228$ with (1) being satisfied for every $x\ge x_0$, but one has to neccesarily assume the Riemann Hypothesis. In a recent paper by A. W. Dudek and D. J. Platt ([2], Theorem 1.2), it has been established that, ramanujan’s Inequality holds true unconditionally for every $x\ge exp\left(9658\right)$. Although this can be considered as an exceptional achievement in this area, efforts of further improvements to this bound are already underway. For instance, Mossinghoff and Trudgian[5] made significant progress in this endeavour, when they established a better estimate as, $x\ge exp\left(9394\right)$. Later on, Platt and Trudgian ([18], cf. Th. 2) together established that, further improvement is indeed possible, and that $x\ge exp\left(3915\right)$. Worth mentioning that, Cully-Hugill and Johnston ([19], cf. Cor. 1.6) literally took it to the next level by obtaining an effective bound for (1) to hold unconditionally as, $x\ge exp\left(3604\right)$. Unsurprisingly, Johnston and Yang ([20], cf. Th. 1.5) outperformed them in claiming the lower bound for such x satisfying Ramanujan’s Inequality to be $exp\left(3361\right)$.

One recent even better result by Axler [6] suggests that, the lower bound for x, namely $exp\left(3361\right)$ can in fact be further improved upto $exp\left(3158.442\right)$ using similar techniques as described in [2], although modifying the error term accordingly adhering to a sharper bound involving $\pi \left(x\right)$ and $Li\left(x\right)$ derived by Fiori, Kadiri, and Swidinsky ([4], cf. Cor. 22).

This paper does indeed adopts a new approach in modifying the esisting estimates for $x_0$ in order for the Ramanujan’s Inequality (cf. Theorem (1)) to hold without imposing any further assumptions on it for every $x\ge x_0$. By utilizing some effective bounds on the Chebyshev’s ϑ-function, the primary intention is to obtain a suitable bound for $\pi \left(x\right)$, and hence eventually come up with a much better estimate for $x_0$ by tinkering with the constants while respecting all the stipulated conditions available to us.

2. An Improved Criterion for Ramanujan’s Inequality

Suppose, we define,

$$\begin{array}{c}\hfill \mathcal{G}\left(x\right):={\left(\pi \left(x\right)\right)}^2-{\displaystyle \frac{ex}{logx}}\pi \left({\displaystyle \frac{x}{e}}\right)\end{array}$$

(2)

A priori using the Prime Number Theorem ([15], cf. Th. 2.1.1), we can in fact assert that [2],

$$\begin{array}{c}\hfill \pi \left(x\right)=x\sum _{k=0}^4{\displaystyle \frac{k!}{{log}^{k+1}x}}+O\left({\displaystyle \frac{x}{{log}^{6}x}}\right)\end{array}$$

(3)

as $x\to \infty$. On the other hand, for the Chebyshev’s ϑ-function having the following definition,

$$\begin{array}{c}\hfill \vartheta \left(x\right):=\sum _{p\le x}logp,\end{array}$$

(4)

we can indeed summarize certain inequalities (cf. [7] and [8]) as follows:

Proposition 1. 

The following holds true for $\vartheta \left(x\right)$:

• $\vartheta \left(x\right)<x$,                           for $x<{10}^8$,
• $|\vartheta \left(x\right)-x|<2.05282\sqrt{x}$,        for $x<{10}^8$,
• $\left|\vartheta \left(x\right)-x\right|<0.0239922{\displaystyle \frac{x}{logx}}$,     for $x\ge 758711$,
• $|\vartheta \left(x\right)-x|<0.0077629{\displaystyle \frac{x}{logx}}$,     for $x\ge exp\left(22\right)$,
• $|\vartheta \left(x\right)-x|<8.072{\displaystyle \frac{x}{{log}^{2}x}}$,          for $x>1$.

Applying these inequalities, we can compute a suitable bound for $\vartheta \left(x\right)$ as follows:

Lemma 1 

(cf. [9]). We shall have the following estimate for $\vartheta \left(x\right)$:

$$\begin{array}{c}\hfill x\left(1-{\displaystyle \frac{2}{3{(logx)}^{1.5}}}\right)<\vartheta \left(x\right)<x\left(1+{\displaystyle \frac{1}{3{(logx)}^{1.5}}}\right),for x\ge 6400.\end{array}$$

(5)

Lemma (1) does in fact enables us deduce a more effective bound for $\pi \left(x\right)$, which’ll prove to be immensely beneficial for us later on.

Theorem 2. 

We shall have the following estimate for $\pi \left(x\right)$ as follows:

$$\begin{array}{c}\hfill {\displaystyle \frac{x}{logx-1+\frac{1}{\sqrt{logx}}}}<\pi \left(x\right)<{\displaystyle \frac{x}{logx-1-\frac{1}{\sqrt{logx}}}},forx\ge 59.\end{array}$$

(6)

We briefly discuss the proof of the Theorem above following the steps as described in [9] for the convenience of our readers.

Proof. 

Applying a well-known inequlity involving $\vartheta \left(x\right)$ and $\pi \left(x\right)$,

$$\begin{array}{c}\hfill \pi \left(x\right)={\displaystyle \frac{\vartheta \left(x\right)}{logx}}+\underset{2}{\overset{x}{\int }}{\displaystyle \frac{\vartheta \left(t\right)}{t{log}^{2}t}}dt\end{array}$$

(7)

and, with the help of (5) in Lemma (1), we get,

$$\begin{array}{c}\hfill \pi \left(x\right)<{\displaystyle \frac{x}{logx}}+{\displaystyle \frac{x}{3{(logx)}^{2.5}}}+\underset{2}{\overset{x}{\int }}{\displaystyle \frac{dt}{{log}^{2}t}}+{\displaystyle \frac{1}{3}}\underset{2}{\overset{x}{\int }}{\displaystyle \frac{dt}{{(logx)}^{3.5}}}\end{array}$$

$$\begin{array}{c}\hfill ={\displaystyle \frac{x}{logx}}\left(1+{\displaystyle \frac{1}{3{(logx)}^{1.5}}}+{\displaystyle \frac{1}{logx}}\right)-{\displaystyle \frac{2}{{log}^{2}2}}+2\underset{2}{\overset{x}{\int }}{\displaystyle \frac{dt}{{log}^{3}t}}+{\displaystyle \frac{1}{3}}\underset{2}{\overset{x}{\int }}{\displaystyle \frac{dt}{{(logt)}^{3.5}}}\end{array}$$

$$\begin{array}{c}\hfill <{\displaystyle \frac{x}{logx}}\left(1+{\displaystyle \frac{1}{3{(logx)}^{1.5}}}+{\displaystyle \frac{1}{logx}}\right)+{\displaystyle \frac{7}{3}}\underset{2}{\overset{x}{\int }}{\displaystyle \frac{dt}{{log}^{3}t}}\end{array}$$

(8)

Moreover, defining the function,

$$\begin{array}{c}\hfill h_1\left(x\right):={\displaystyle \frac{2}{3}}.{\displaystyle \frac{x}{{(logx)}^{2.5}}}-{\displaystyle \frac{7}{3}}{\displaystyle \frac{dt}{{log}^{3}t}},\mathrm{for} x\ge exp\left(18.25\right)\end{array}$$

(9)

We can observe that, $h_1^{\prime }\left(x\right)>0$, implying that, $h_1$ is increasing. Now, for every convex function $u:[a,b]\to \mathbb{R}$, where, $a<b,a,b\in {\mathbb{R}}_{>0}$, we have,

$$\begin{array}{c}\hfill \underset{a}{\overset{b}{\int }}u\left(x\right)dx\le {\displaystyle \frac{b-a}{n}}\left(u\left(a\right)+u\left(b\right)+\sum _{k=1}^{n-1}u\left(a+k{\displaystyle \frac{b-a}{n}}\right)\right).\end{array}$$

(10)

Thus, choosing $u\left(x\right):={\displaystyle \frac{1}{{log}^{3}x}}$ and $n={10}^5$ and using (10) on each of the intervals $[2,e]$, $[e,e^2]$, ......, $[e^{17},e^{18}]$ and $[e^{18},e^{18.25}]$ yields,

$$\begin{array}{c}\hfill \underset{2}{\overset{exp\left(18.25\right)}{\int }}{\displaystyle \frac{dt}{{log}^{3}t}}<16870.\end{array}$$

Furthermore, one can also verify using MATHEMATICA that,

$$\begin{array}{c}\hfill h_1(exp\left(18.25\right))>{\displaystyle \frac{1}{3}}(118507-118090)>0.\end{array}$$

Therefore, for every $x\ge exp\left(18.25\right)$, we must have from (8),

$$\begin{array}{c}\hfill \pi \left(x\right)<{\displaystyle \frac{x}{logx}}\left(1+{\displaystyle \frac{1}{3{(logx)}^{1.5}}}+{\displaystyle \frac{1}{logx}}\right)<{\displaystyle \frac{x}{logx-1-\frac{1}{\sqrt{logx}}}}\end{array}$$

(11)

Again, for $x\le exp\left(18.25\right)<{10}^8$, we apply $\left(1\right)$ in Proposition (1) to derive,

$$\begin{array}{c}\hfill \pi \left(x\right)={\displaystyle \frac{\vartheta \left(x\right)}{logx}}+\underset{2}{\overset{x}{\int }}{\displaystyle \frac{\vartheta \left(x\right)}{t{log}^{2}t}}dt<{\displaystyle \frac{x}{logx}}+\underset{2}{\overset{x}{\int }}{\displaystyle \frac{dt}{{log}^{2}t}}\end{array}$$

$$\begin{array}{c}\hfill ={\displaystyle \frac{x}{logx}}\left(1+{\displaystyle \frac{1}{logx}}\right)-{\displaystyle \frac{2}{{log}^{2}2}}+2\underset{2}{\overset{x}{\int }}{\displaystyle \frac{dt}{{log}^{3}t}}.\end{array}$$

Furthermore, for $4000\le x<{10}^8$, taking the function,

$$\begin{array}{c}\hfill h_2\left(x\right):={\displaystyle \frac{x}{{(logx)}^{2.5}}}-2\underset{2}{\overset{x}{\int }}{\displaystyle \frac{dt}{{log}^{3}t}}+{\displaystyle \frac{2}{{log}^{2}2}}.\end{array}$$

(12)

We can indeed verify that, $h_2^{\prime }\left(x\right)>0$, implying $h_2$ is an increasing function. Similarly, with the help of MATHEMATICA, we can compute the sign of $h_2$ as follows,

$$\begin{array}{c}\hfill h_2(exp\left(11\right))>149-2\underset{2}{\overset{exp\left(11\right)}{\int }}{\displaystyle \frac{dt}{{log}^{3}t}}>149-140>0.\end{array}$$

In summary, thus for $exp\left(11\right)\le x<{10}^8$,

$$\begin{array}{c}\hfill \pi \left(x\right)<{\displaystyle \frac{x}{logx}}\left(1+{\displaystyle \frac{1}{logx}}+{\displaystyle \frac{1}{{(logx)}^{1.5}}}\right)<{\displaystyle \frac{x}{logx-1-\frac{1}{\sqrt{logx}}}}.\end{array}$$

(13)

In addition to the above, it is important to note that, for $x\ge 6$, the denominator, $logx-1-{\displaystyle \frac{1}{\sqrt{logx}}}>0$. Which means that, for $6\le x\le exp\left(11\right)$, we need to establish,

$$\begin{array}{c}\hfill H\left(x\right):={\displaystyle \frac{x}{\pi \left(x\right)}}+1+{(logx)}^{-0.5}-logx>0.\end{array}$$

(14)

Assuming $p_n$ to be the $n^{th}$ prime, it can be observed that, H is in fact increasing in $\left[p_n,p_{n+1}\right)$, thus it only needs to be proven that, $H\left(p_n\right)>0$.

For $p_n<exp\left(11\right)$, we have the inequality ${\displaystyle \frac{1}{\sqrt{log{p}_n}}}>0.3$, which reduces our computation to verifying,

$$\begin{array}{c}\hfill {\displaystyle \frac{p_n}{n}}-log{p}_n>-1.3\end{array}$$

for every $7\le p_n\le exp\left(11\right)$, which can be achieved using MATHEMATICA.

In order to establish the lower bound of $\pi \left(x\right)$ as claimed in (6), we shall be needing $\left(1\right)$ in Proposition (1) and (5) in Lemma (1) under the condition that, $x\ge 6400$. Hence,

$$\begin{array}{c}\hfill \pi \left(x\right)-\pi \left(6400\right)={\displaystyle \frac{\vartheta \left(x\right)}{logx}}-{\displaystyle \frac{\vartheta \left(6400\right)}{log\left(6400\right)}}+\underset{6400}{\overset{x}{\int }}{\displaystyle \frac{\vartheta \left(t\right)}{t{log}^{2}t}}dt.\end{array}$$

(15)

Rigorous computations does yield, $\pi \left(6400\right)=834$, and, ${\displaystyle \frac{\vartheta \left(6400\right)}{log\left(6400\right)}}<{\displaystyle \frac{6400}{log\left(6400\right)}}<731$. Thus, (15) further reduces to,

$$\begin{array}{c}\hfill \pi \left(x\right)>103+{\displaystyle \frac{\vartheta \left(x\right)}{x}}+\underset{6400}{\overset{x}{\int }}{\displaystyle \frac{\vartheta \left(t\right)}{t{log}^{2}t}}dt.\end{array}$$

Using the lower bound of $\vartheta \left(x\right)$ as in (5) of Lemma (1) gives,

$$\begin{array}{c}\hfill \pi \left(x\right)>103+{\displaystyle \frac{x}{logx}}-{\displaystyle \frac{2x}{3{log}^{2.5}x}}+{\displaystyle \frac{x}{{log}^{2}x}}-{\displaystyle \frac{6400}{{log}^{2}6400}}+2\underset{6400}{\overset{x}{\int }}{\displaystyle \frac{dt}{{log}^{3}t}}-{\displaystyle \frac{2}{3}}\underset{6400}{\overset{x}{\int }}{\displaystyle \frac{dt}{{log}^{3.5}t}}\end{array}$$

$$\begin{array}{c}\hfill >{\displaystyle \frac{x}{logx}}\left(1+{\displaystyle \frac{1}{logx}}-{\displaystyle \frac{2}{3{log}^{1.5}x}}\right)>{\displaystyle \frac{x}{logx-1+\frac{1}{\sqrt{logx}}}}\end{array}$$

Setting $v={(logx)}^{-0.5}$, we can assert that, the above inequality holds true for, $2v^3-5v^2+3v-1<0$, implying, $v(1-v)(3-2v)\le {\displaystyle \frac{(3-v)}{4}}<1$. Hence, it can be confirmed that, the statement (6) holds true for $x\ge 6400$.

Furthermore, for $x<6400$, we intend on showing that,

$$\begin{array}{c}\hfill \beta \left(x\right):=-{\displaystyle \frac{x}{\pi \left(x\right)}}+logx-1+{\displaystyle \frac{1}{\sqrt{logx}}}>0.\end{array}$$

(16)

Assuming similarly that, $p_n$ denotes the $n^{th}$ prime, one can observe that, the function $\beta \left(x\right)$ is indeed decreasing on $\left[p_n,p_{n+1}\right)$. Hence, it only suffices to check for the values at $p_n-1$. Now, $p_n\le 6400$ implies, ${(log(p_n-1))}^{-0.5}>0.337$, and thus, it only is needed to be checked that,

$$\begin{array}{c}\hfill {\displaystyle \frac{log(p_n-1)}{p_n-1}}-{\displaystyle \frac{p_n-1}{n-1}}>0.663\end{array}$$

(17)

Utilizing proper coding in MATHEMATICA gives us, $n\ge 36$ in order for (17) to satisfy. Therefore, we can further verify that, (6) holds for $x\ge 59$, and the proof is complete.    □

Significantly, Karanikolov [10] cited one of the applications of (6) which says that for $\alpha \ge e^{1/4}$ and, $x\ge 364$, we must have,

$$\begin{array}{c}\hfill \pi \left(\alpha x\right)<\alpha \pi \left(x\right).\end{array}$$

(18)

Although, a more effective version of (18) states (cf. Theorem 2 [9]) the following.

Proposition 2. 

(18) holds true for every $\alpha >1$ and, $x>exp\left(4{(log\alpha )}^{-2}\right)$,

Proof. 

We utilize (6) in theorem (2) for $\alpha x\ge 6$. Thus,

$$\begin{array}{c}\hfill {\displaystyle \frac{\alpha x}{log\alpha x-1+\frac{1}{\sqrt{log\alpha x}}}}<\pi \left(\alpha x\right)<{\displaystyle \frac{\alpha x}{log\alpha x-1-\frac{1}{\sqrt{log\alpha x}}}}\end{array}$$

and,

$$\begin{array}{c}\hfill {\displaystyle \frac{\alpha x}{logx-1+\frac{1}{\sqrt{logx}}}}<\alpha \pi \left(x\right)<{\displaystyle \frac{\alpha x}{logx-1-\frac{1}{\sqrt{logx}}}}\end{array}$$

for every $x\ge 59$. Now, assuming $x\ge exp\left(4{(log\alpha )}^{-2}\right)$, we can deduce that,

$$\begin{array}{c}\hfill log\alpha >{(log\alpha x)}^{-0.5}+{(logx)}^{-0.5}\end{array}$$

which is all that we’re required to show. This completes the proof.    □

As for another application of (6), we must mention the work of Udrescu [11], where it was claimed that, if $0<ϵ\le 1$, then,

$$\begin{array}{c}\hfill \pi (x+y)<\pi \left(x\right)+\pi \left(y\right),\forall ϵx\le y\le x.\end{array}$$

(19)

Again, further progress have in fact been made in order to improve the result (19). One such notable work in this regard has been done by Panaitopol [9].

Lemma 2 

(19) is satisfied under additional condition, $x\ge exp\left(9{ϵ}^{-2}\right)$, where, $ϵ\in \left(0,1\right]$.

We shall be using all the above derivations in order to obtain a much improved bound for $x_0$ such that, $\mathcal{G}\left(x\right)<0$ unconditionally for every $x\ge x_0$.

Choose some $a>1$ such that, $e-a>a>1$ as well. Hence,

$$\begin{array}{c}\hfill \pi \left(x\right)=\pi \left(e.{\displaystyle \frac{x}{e}}\right)=\pi \left(a.{\displaystyle \frac{x}{e}}+(e-a).{\displaystyle \frac{x}{e}}\right)\end{array}$$

(20)

Using (19) by taking, $ϵ={\displaystyle \frac{a}{e-a}}<1$ as per our construction yields,

$$\begin{array}{c}\hfill \pi \left(x\right)<\pi \left(a.{\displaystyle \frac{x}{e}}\right)+\pi \left((e-a).{\displaystyle \frac{x}{e}}\right)\end{array}$$

(21)

for every $x\ge {\displaystyle \frac{e}{e-a}}.exp\left(9.{\left({\displaystyle \frac{a}{e-a}}\right)}^{-2}\right)$. Furthermore, by our selection of a, we can in fact utilize Proposition (2) again in order to derive the following estimates,

$$\begin{array}{c}\hfill \pi \left(a.{\displaystyle \frac{x}{e}}\right)<a.\pi \left({\displaystyle \frac{x}{e}}\right),\forall x>exp\left(4{(loga)}^{-2}+1\right)\end{array}$$

(22)

and,

$$\begin{array}{c}\hfill \pi \left((e-a).{\displaystyle \frac{x}{e}}\right)<(e-a).\pi \left({\displaystyle \frac{x}{e}}\right),\forall x>exp\left(4{(log(e-a))}^{-2}+1\right)\end{array}$$

(23)

Therefore, combining (20), (21), (22) and (23), we obtain,

$$\begin{array}{c}\hfill \pi \left(x\right)<a.\pi \left({\displaystyle \frac{x}{e}}\right)+(e-a).\pi \left({\displaystyle \frac{x}{e}}\right)=e.\pi \left({\displaystyle \frac{x}{e}}\right)\end{array}$$

(24)

for every such,

$$\begin{array}{c}\hfill x\ge max\left\{{\displaystyle \frac{e}{e-a}}.exp\left(9.{\left({\displaystyle \frac{a}{e-a}}\right)}^{-2}\right),exp\left(4{(loga)}^{-2}+1\right),exp\left(4{(log(e-a))}^{-2}+1\right)\right\}\end{array}$$

(25)

For our convenience, we consider, $a=1.3003232>1$.

Thus, we can verify, $ϵ=0.9170389<1$. Subsequently, we conclude that, (24) is satisfied for every,

$$\begin{array}{c}\hfill x\ge max\left\{1.917038924exp\left(10.7020504\right),exp\left(59\right),exp\left(33.799421\right)\right\}\end{array}$$

(26)

In summary, we have,

$$\begin{array}{c}\hfill \pi \left(x\right)<e.\pi \left({\displaystyle \frac{x}{e}}\right),\forall x\ge exp\left(59\right).\end{array}$$

(27)

On the other hand, (3) gives us,

$$\begin{array}{c}\hfill \pi \left(x\right)>{\displaystyle \frac{x}{logx}}\end{array}$$

(28)

for sufficiently large values of x. Finally, combining (27) and (28), we get from (2),

$$\begin{array}{c}\hfill \mathcal{G}\left(x\right)={\left(\pi \left(x\right)\right)}^2+\left({\displaystyle \frac{x}{logx}}\right).\left(-e.\pi \left({\displaystyle \frac{x}{e}}\right)\right)<{\left(\pi \left(x\right)\right)}^2+\pi \left(x\right).(-\pi \left(x\right))=0.\end{array}$$

(29)

and this is valid unconditionally for every $x\ge exp\left(59\right)$. Therefore, we have our $x_0=exp\left(59\right)$ as desired in order for the Ramanujan’s Inequality to hold without any further assumptions.

3. Numerical Estimates for $\mathcal{G}\left(x\right)$

We can indeed verify our claim using programming tools such as MATHEMATICA for example. The numerical data1 from the Table 1 and the plot Figure 1 representing $log(-\mathcal{G}(x\left)\right)$ against $logx$ for $x\in [exp(59),exp(3159\left)\right]$ clearly establishes that, $\mathcal{G}$ is indeed monotone decreasing in the said interval, and also is strictly negative. It only suffices to check until $exp\left(3159\right)$, as the result has been unconditionally proven for $x\ge exp\left(3158.442\right)$ by Axler [6].

4. Future Research Prospects

In summary, we’ve utilized specific order estimates for the Prime Counting Function $\pi \left(x\right)$ in addition to several explicit bounds involving Chebyshev’s ϑ-function, $\vartheta \left(x\right)$, a priori with the help of the Prime Number Theorem in order to conjure up an improved bound for the famous Ramanujan’s Inequality. Although, it’ll surely be interesting to observe whether it’s at all feasible to apply any other techniques for this purpose.

On the other hand, one can surely work on some modifications of Ramanujan’s Inequality For instance, Hassani studied (1) extensively for different cases [3], and eventually claimed that, the inequality does in fact reverses if one can replace e by some $\alpha$ satifying, $0<\alpha <e$, although it retains the same sign for every $\alpha \ge e$.

In addition to above, it is very much possible to come up with certain generalizations of Theorem (1). In this context, we can study Hassani’s stellar effort in this area where, he apparently increased the power of $\pi \left(x\right)$ from 2 upto $2^n$ and provided us with this wonderful inequality stating that for sufficiently large values of x [16],

$$\begin{array}{c}\hfill {\left(\pi \left(x\right)\right)}^{2^n}<{\displaystyle \frac{e^n}{\prod _{k=1}^n{\left(1-\frac{k-1}{logx}\right)}^{2^{n-k}}}}{\left({\displaystyle \frac{x}{logx}}\right)}^{2^n-1}\pi \left({\displaystyle \frac{x}{e^n}}\right)\end{array}$$

Finally, and most importantly, we can choose to broaden our horizon, and proceed towards studying the prime counting function in much more detail in order to establish other results analogous to Theorem (1), or even study some specific polynomial functions in $\pi \left(x\right)$ and also their powers if possible. One such example which can be found in [17] eventually proves that, for sufficiently large values of x,

$$\begin{array}{c}\hfill {\displaystyle \frac{3ex}{logx}}{\left(\pi \left({\displaystyle \frac{x}{e}}\right)\right)}^{3^n-1}<{\left(\pi \left(x\right)\right)}^{3^n}+{\displaystyle \frac{3e^{2}x}{{(logx)}^2}}{\left(\pi \left({\displaystyle \frac{x}{e^2}}\right)\right)}^{3^n-2},n>1\end{array}$$

Whereas, significantly the inequality reverses for the specific case when, $n=1$ (Cubic Polynomial Inequality) (cf. Theorem $3.1.1$ [17]).

Hopefully, further research in this context might lead the future researchers to resolve some of the unsolved mysteries involving prime numbers, or even solve some of the unsolved problems surrounding the iconic field of Number Theory.