20
$\begingroup$

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 $\frac{2}{qr}$ .

$\endgroup$
  • 1
    $\begingroup$ Where did you find this claim? $(+1)$ $\endgroup$ – Mr Pie Apr 21 '18 at 9:06
  • 4
    $\begingroup$ @user477343 I made it by myself... $\endgroup$ – Peđa Terzić Apr 21 '18 at 9:11
  • 2
    $\begingroup$ Wow... brilliant :) $\endgroup$ – Mr Pie Apr 21 '18 at 9:18
  • 4
    $\begingroup$ @gnasher729 : The claim says three consecutive primes. $\endgroup$ – Nilotpal Kanti Sinha Apr 22 '18 at 5:17
  • 1
    $\begingroup$ @gnasher729 $$\frac{1}{101^2}<\frac{1}{103^2}+\frac{1}{107^2}$$ as $(101,103,107)$ is a triplet of three consecutive primes. $\endgroup$ – George N. Missailidis Apr 23 '18 at 12:34
14
$\begingroup$

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 $$ \frac{1}{q^2}+\frac{1}{r^2}>\frac{1}{a^2\,p^2}+\frac{1}{a^4\,p^2}=\frac{1}{p^2}. $$

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

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 $$\frac{1}{\big(2(p'+q'+r')-q-r\big)^2} < \frac{1}{q^2} + \frac{1}{r^2}.$$

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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 $p_{\epsilon_a}$

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.