On some conjectures about an inequality involving different arithmetic functions related to prime numbers I've written the following puzzle about prime numbers. This exercise is thus a curiosity/miscellany about the distribution of prime number $p_k$, that I wondered when I was playing with different arithmetic functions. Thus maybe the inequality is artificious.
For integers $n\geq 1$ in this post we denote the $n$th prime number as $p_n$, $\pi(n)$ is the prime-counting function. And $\operatorname{rad}(n)$ denotes the product of distinct pimes dividing an integer $n>1$, with the definition $\operatorname{rad}(1)=1$, that is the so-called radical of an integer (see this Wikipedia).
I've consider integers, or equivalently the indexes of those prime numbers $p_n$, satisfying for some $\epsilon>0$
$$n^{\frac{1}{2}+\epsilon}\cdot p_n<\frac{\pi(n)}{\pi(2n)}\cdot\operatorname{rad}\left(\sum_{k=1}^n k\right).\tag{1}$$
Here of course we can write $\frac{n(n+1)}{2}$ instead of $\sum_{k=1}^n k$.

Question. Prove or refute the following conjectures:
A) When $\epsilon =0$ there are infinitely many integers $n\geq 1$ (or as you want to see it, the subscripts $n$'s of primes $p_n$) satisfying the inequality $(1)$.
B) If $n\geq 1$ is a solution of $(1)$ then our real $0<\epsilon$ satisfies $\epsilon<\frac{1}{2}$.
Many thanks.

In summary I am asking how to attack these exercises involving the inequality. I believe thus that there are solutions only for $0<\epsilon<\frac{1}{2}$. About B) of course, if you find counterexamples you can solve the Question invoking those. If A) is very difficult to solve, I should to accept some answer showing what work can be done or a remarkable heuristic/reasoning. Also computational evidence is welcome as attached to your answer or in comments.
 A: I'm afraid the discussion of prime numbers here and the various counting functions is largely a red herring; this is a pure matter of asymptotics and squarefreeness.  Note that since $\gcd(n,n+1)=1$, $\mathrm{rad}(n(n+1)/2)$ is either $\mathrm{rad}(n/2)\mathrm{rad}(n+1)$ or $\mathrm{rad}(n)\mathrm{rad}((n+1)/2)$. But $\mathrm{rad}(n)=n$ if $n$ is squarefree, and it's known that there are infinitely many (in fact, a positive proportion) of consecutive squarefree numbers; see this MathOverflow question for some discussion and references. Thus, infinitely often you'll have $\mathrm{rad(n(n+1)/2)}=n(n+1)/2$.  From here, the rest is easy asymptotics: as you noted yourself, $\frac{\pi(n)}{\pi(2n)}\to\frac12$, so your RHS is $\Theta(n^2)$ on a set of positive density; since your LHS is $\Theta(n^{(3/2)+\epsilon}\log n)$ then the inequality will hold 'asymptotically' (and therefore infinitely often) for any $\epsilon\lt\frac12$.
OTOH, when $\epsilon=\frac12$ the LHS is $\Theta(n^2\log n)$ and the RHS is (everywhere, not just on a positive-density set) $O(n^2)$ so there can be at most finitely many counterexamples; it shouldn't be hard to find explicit bounds for all of the quantities involved, and from there put a bound on the size of any counterexamples that's within range of easy hand (or computer-aided) checking.
