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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.

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  • $\begingroup$ 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. $\endgroup$ – user142929 Aug 7 at 6:25
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    $\begingroup$ 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. $\endgroup$ – JoshuaZ Aug 8 at 15:50
  • $\begingroup$ 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. $\endgroup$ – user142929 Aug 8 at 21:56
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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 $y$.)

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  • $\begingroup$ Many thanks, I am going to study it, seems very elegant. $\endgroup$ – user142929 Aug 7 at 20:14
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    $\begingroup$ @user142929 See update. $\endgroup$ – Stopple Aug 8 at 15:26

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