# A simple inequality that arises from the exact form for the prime-counting function and the second Hardy–Littlewood conjecture

The germ of this post arises when I was trying to think what terms of the so-called exact form of the prime-counting function satisfy an inequality of the form $$\text{term}(x+y)\leq \text{term}(x)+\text{term}(y)$$, or of the form $$\text{term}(x+y)\leq \text{term}(x)+\text{term}(y)$$.

Is in the literature several problems related to the prime-counting function $$\pi(x)$$, one of the most famous is the Riemann hypothesis. Riemann provide us a formula, called the exact form, that you can see in the section Exact form of the Wikipedia Prime counting function. But is in the literature other unsolved problem, a less famous problem the Second Hardy–Littlewood conjecture, see the corresponding Wikipedia.

My belief is the following conjecture should be easy to get, since I believe that the mistery of these unsolved problems does not lie in the term that I evoke in the following conjecture, any case I believe that this question is interesting for this site, and I'm curious to know how you analyze the inequality.

Conjecture. For real numbers $$x\geq 2$$ and $$y\geq 2$$ the following inequality holds

$$\begin{multline} \frac{1}{\pi}\left(\arctan\left(\frac{\pi}{\log x}\right)+\arctan\left(\frac{\pi}{\log y}\right)-\arctan\left(\frac{\pi}{\log (x+y)}\right)\right) \\ \leq\frac{1}{\log x}+\frac{1}{\log y}-\frac{1}{\log (x+y)}. \end{multline}$$

Question. Can you prove previous conjecture? Many thanks.

I know methods to solve inequalities more easy than this.

• I don't know if the best tag is (analytic-number-theory), or there is a tag more suitable than this, if this question is welcome an interesting, feel free to add/remove those tags more suitable. Feel free to comment if it is possible to improve the mathematical content of my post. – user142929 Aug 7 at 6:25
• Note that there's an easy proof for when x=y. If one sets f(x) to be the RHS - LHS in your above proposed inequality, then it is not too hard to see that the limit as x goes to infinity is 0, and to check that f'(x) is negative except for a pole a little smaller than 2. If f(x) was negative for some x in that range, then one would have a contradiction. – JoshuaZ Aug 8 at 15:50
• Many thanks to you @JoshuaZ and Stopple for yours contributions, usually in my home I am going to take notes as remarks in a notebook. – user142929 Aug 8 at 21:56

Suggestion: We have the series expansion $$\arctan(\pi/z)/\pi=\sum_{k=0}^\infty (-1)^k \frac{\pi^{2k}}{(2k+1)z^{2k+1}}$$ with $$z$$ respectively $$\log(x)$$, $$\log(y)$$, and $$\log(x+y)$$. This is an alternating series with terms tending to $$0$$. Since the partial sums alternately overestimates and underestimates the sum, it would suffice to prove the simpler inequality $$\frac{\pi^2}{3\log(x+y)^3}\le\frac{\pi^2}{3\log(x)^3}+\frac{\pi^2}{3\log(y)^3}-\frac{\pi^4}{5\log(x)^5}-\frac{\pi^4}{5\log(y)^5}.$$
Update: For $$y\ge 12$$, $$\frac{\pi^2}{3\log(y)^3}-\frac{\pi^4}{5\log(y)^5}>0,$$ and for any fixed $$x$$, as $$y\to\infty$$ $$\frac{\pi^2}{3\log(x+y)^3}\to 0,$$ so certainly as $$y\to\infty$$ $$\frac{\pi^2}{3\log(x+y)^3}\le\frac{\pi^2}{3\log(x)^3}-\frac{\pi^4}{5\log(x)^5}.$$ Thus the desired inequality holds in the region above some curve, $$y\ge y(x)$$ (and the same reversing the roles of $$x$$ and $$y$$.)