An inequality regarding the sum of primes Let $p_n$ be the $n$th prime.
For integers $k \ge 3, c \ge 1$, does it follow that:
$$\sum_{k \le i < k+c}(p_i - 2c) > 0$$
If not, is there a minimum value of $k$ where it is true?
I know that it is not true for $k=2, c=2$ or $k=2, c=3$ since:


*

*$(3 - 4) + (5 - 4) = 0$

*$(3 - 6) + (5 - 6) + (7 - 6) = -3$
The base case is true for $k \ge 3$ since $5 > 2$.
Is there a straight forward way to prove this?  Is this a very difficult question that does not have a known answer?

Edit: Here are some details for $c\ge 13$
I found this paper which suggests for very large $x$ that if $S(x)$ is the sum of primes less than $x$, then:
$$S(x) > \frac{x^2}{2\log x} + \frac{x^2}{4\log^2 x} + \frac{x^2}{4\log^3 x} + \frac{1.2x^2}{8\log^4 x}$$
Massias and Robin showed that for $n \ge 13$:
$$p_n \le n\left(\log n + \log \log n - 1 + \frac{1.8\log\log n}{\log n}\right)$$
So for $c \ge 13$, the $\dfrac{\text{sum of primes}}{\text{count of primes}}$ is greater than:
$\dfrac{S(u)}{c}$
where $u = c\left(\log c + \log \log c - 1 + \frac{1.8\log\log c}{\log c}\right)$
 A: Note that if for a specific $c$ your inequality is true for any particular $k$, it's also true for any larger $k$ as the value on the left would be adding the same number of primes, but with each of them being larger, so the left hand side value would increase. Thus, you only really need to check for $k = 3$.
Your statement is always true for sufficiently large $c$. To see this, note that
$$\begin{equation}\begin{aligned}
\sum_{i=3}^{c+2}(p_i - 2c) & \gt 0 \\
\sum_{i=3}^{c+2}p_i - \sum_{i=3}^{c+2}(2c) & \gt 0 \\
\sum_{i=3}^{c+2}p_i - 2c((c+2) - (3) + 1) & \gt 0 \\
\sum_{i=3}^{c+2}p_i & \gt 2c^2
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
The Prime number theorem states that
$$p_n \sim n\log(n) \tag{2}\label{eq2A}$$
Now, $\log(n) = 4 \implies n \approx 55$, and $\log(n)$ is an unbounded strictly increasing function, so for large enough $c$, you have that
$$\begin{equation}\begin{aligned}
\sum_{i=3}^{c+2}p_i & \gt \sum_{i=3}^{c+2}4i \\
& = 4\left(\frac{(c+2)(c+3)}{2} - 3\right) \\
& = 2(c^2 + 5c) \\
& \gt 2c^2
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Thus, you just need to check for $k = 3$ and values of $c$ up to some reasonably large value. This is even just a few hundred, but you may wish to go to something like a thousand or even $1,000,000$ to be absolutely sure. In addition to this, or alternatively if you want to check by hand instead, you can use some of the expressions in the Approximations for the $n$th prime number section of Wikipedia's PNT article, such as that
$$p_n \gt n(\log n + \log \log n - 1), \; n \ge 6 \tag{4}\label{eq4A}$$
This means, for example, that $p_n \gt n\log n$ where $\log \log n \gt 1$, which occurs for $n \ge 16$.
