<s>[INST] An inequality involving three consecutive primes 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 $\frac{2}{qr}$  .
 [/INST] 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 $$\frac{1}{\big(2(p'+q'+r')-q-r\big)^2} < \frac{1}{q^2} + \frac{1}{r^2}.$$
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