An inequality holding for odd primes

Question

Let $P_{\Sigma}(n)$ be the sum of all prime factors of $n$. Then there seems to be an inequality
$$p\ge P_{\Sigma}(rad(p^2-1))$$ for all odd primes $p$. Tested for the first million primes. Is this a known fact?

Answer 1

Let $P(n):= P_{\Sigma}(rad(n))$ to simplify the notation below. We want to show $P(p^2-1)\leq p$ for all odd primes $p$.

Notice that $\gcd(p-1,p+1)=2$ since $p$ is odd, and at least one of $p-1$ and $p+1$ will be divisible by $4$, hence we do not lose any primes by dividing both terms by $2$, and this also makes them coprime. So

$$P(p^2-1)=P((p-1)(p+1))=P\Bigl(\frac{p-1}{2}\frac{p+1}{2}\Bigr)=P\Bigl(\frac{p-1}{2}\Bigr)+P\Bigl(\frac{p+1}{2}\Bigr).(*)$$

The remaining factors are small, the following upper bound on $P(n)$ will help:

Lemma. $P(n)\leq n$ for all $n \geq 2$.

Proof. We can show a stronger statement $P_{\Sigma}(n) \leq n$, i.e. the inequality holds even when multiplicities are counted. Maybe there is more elementary method, but following strong induction works. First notice that $P_{\Sigma}(2)=2 \leq 2$. Then assume $P_{\Sigma}(k)\leq k$ for $k=2,3,\dots,n$ and we want to show $P_{\Sigma}(n+1)\leq n+1$. If $n+1=p^e$ is a prime power, this follows directly as $P_{\Sigma}(p^e)=pe\leq p^e$ (the last inequality can itself be shown by induction on $e$). If $n+1$ is not a prime power, we can write $n+1=ab$ where $a,b\geq 2$ and $\gcd(a,b)=1$. But then $P_{\Sigma}(n+1)=P_{\Sigma}(ab)=P_{\Sigma}(a)+P_{\Sigma}(b)\leq a+b\leq ab=n+1$. $\square$

Using the lemma on $(*)$ gives the desired claim $$ P(p^2-1)\leq\frac{p-1}{2}+\frac{p+1}{2}=p. $$ Note: We have not used that $p$ is a prime at all, so the claim holds for all $p\in \mathbb{N}$ odd.