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Can you provide a proof or a counterexample to the following claim :

Let $p,q,r$ be three consecutive prime numbers such that $p\ge 11 $ and $p<q<r$ , then $\frac{1}{p^2}< \frac{1}{q^2} + \frac{1}{r^2}$ .

I have tested this claim up to $10^{10}$ .

For $p>5$ we get $\pi(2p)-\pi(p) \ge 2$ , a result by Ramanujan . This means that $q<2p$ and $r<2p$ , so $\frac{1}{2p}<\frac{1}{q}$ and $\frac{1}{2p}<\frac{1}{r}$ which implies $\frac{1}{p} < \frac{1}{q} + \frac{1}{r}$ . If we square both sides of inequality we get $\frac{1}{p^2} < \frac{1}{q^2} + \frac{2}{qr} + \frac{1}{r^2}$ . Now , I don't know how to rule out term $\frac{2}{qr}$ .

Martin Sleziak
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Pedja
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    Where did you find this claim? $(+1)$ – Mr Pie Apr 21 '18 at 09:06
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    @user477343 I made it by myself... – Pedja Apr 21 '18 at 09:11
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    Wow... brilliant :) – Mr Pie Apr 21 '18 at 09:18
  • Let me tell you what I'm trying to explore this week, different versions/variations of the so-called Firoozbakht's conjecture. I try rewrite the inequality showed in Firoozbakht's conjecture for different arithmetic functions. My problem is that I can not to find one (from my experiments and knowledges) with a good mathematical content. Maybe a variation using Ramanujan primes instead of prime numbers has mathematical meaning. I say it if you want to explore it. Isn't required a response and good week. –  Apr 21 '18 at 09:36
  • @user477343 To be honest relation $1/p < 1/q+1/r$ was already known...:-) – Pedja Apr 21 '18 at 15:51
  • Your result by Ramanujan doesn't prove the claim, for example 1/101^2 < 1/197^2 + 1/199^2 is false. You need two primes between p and 1.4142p. – gnasher729 Apr 21 '18 at 19:05
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    @gnasher729 : The claim says three consecutive primes. – Nilotpal Sinha Apr 22 '18 at 05:17
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    @gnasher729 $$\frac{1}{101^2}<\frac{1}{103^2}+\frac{1}{107^2}$$ as $(101,103,107)$ is a triplet of three consecutive primes. – George N. Missailidis Apr 23 '18 at 12:34

3 Answers3

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The inequality holds for all $p$ large enough. Let $a>1$ be such that $a^{-2}+a^{-4}=1$ and $p_n$ be the $n$-th prime. By the Prime Number Theorem there is an $N$ such that $p_{n+1}<a\,p_n$ for all $n\ge N$.If $p\ge p_N$, then $q<a\,p$ and $r<a\,q<a^2\,p$ and $$ \frac{1}{q^2}+\frac{1}{r^2}>\frac{1}{a^2\,p^2}+\frac{1}{a^4\,p^2}=\frac{1}{p^2}. $$

Julián Aguirre
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This is a comment as opposed to an answer

All primes $p_{n+1} < 2p_n$. Let $p'$ be the prime before $p$, $q'$ be the prime before $q$, and $r'$ be the prime before $r$. Then, $$p+q+r< 2(p' + q' + r')$$ So $$\frac{1}{p^2} < \frac{1}{\big(2(p' + q' + r') - q - r\big)^2}$$ Also, $$\frac{1}{p^2} < \frac{1}{q^2} + \frac{2}{qr} + \frac{1}{r^2} = \left(\frac{1}{q} + \frac{1}{r}\right)^2.$$ Consider $$\frac{1}{\big(2(p' + q' + r') - q - r\big)^2} < \left(\frac{1}{q} + \frac{1}{r}\right)^2$$ then multiplying both sides by the denominator, subtracting $1$ from both sides, and then factoring, we get $$0 < \left(\left(\frac{1}{q} + \frac{1}{r}\right)\left(2(p'+q'+r')-q-r\right) + 1\right)\left(\left(\frac{1}{q} + \frac{1}{r}\right)\left(2(p'+q'+r')-q-r\right) - 1\right)$$ Which is true since primes are always positive, the entire inequality is literally just a bunch of multiplication, and if $p' = 2$, $q' = 3$ and $r' = 5$ then this inequality holds.

Your conjecture would be thus true if $$\frac{1}{\big(2(p'+q'+r')-q-r\big)^2} < \frac{1}{q^2} + \frac{1}{r^2}.$$

Mr Pie
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Here is a stronger form of the above inequality. The prime number theorem implies that for every $\epsilon$, there exists a prime $p_{\epsilon}$ such that for all $p > p_{\epsilon}$, we have $1 - \epsilon < \frac{p}{q}, \frac{q}{r} < 1$. Hence for all positive real $a$ and all primes greater than some $p_{\epsilon_a}$

$$ \frac{2 - \epsilon_a}{p^a} < \frac{1}{q^a} + \frac{1}{r^a} < \frac{2}{p^a} $$

Nilotpal Sinha
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