By specialization of this formula, here in PROBLEMA 36, page 453 (in spanish), taking $\frac{1}{x_i}$ as the ith prime number we've (with at least two summands) $$ \left( \sum_{k=1}^{n} p_{k} \right)^{2} \geq\frac{2n}{n-1} \sum_{k=1}^{n-1} p_{k} \sum_{l=k+1}^{n} p_{l}, $$ and since $$ \sum_{k=1}^{n} p_{k}=p_n\pi(p_n)- \sum_{k=1}^{p_n-1} \pi(k) =np_n-\frac{1}{2}\frac{p^2_n}{\log p_n} +O \left( \frac{p_n^2}{\log^2p_n} \right) $$ then $$\sum_{k=1}^{n-1} p_{k} \sum_{l=k+1}^{n} p_{l}=\frac{n-1}{2n} \left( np_n-\frac{1}{2}\frac{p_n^2}{\log p_n} +O \left( \frac{p_n^2}{\log^2p_n} \right) \right)^{2}. $$
Question. Is it right? Can you improve it or do a right simplification? If you can improve the computations for the behaviour of $$\sum_{k=1}^{n-1} p_{k} \sum_{l=k+1}^{n} p_{l}$$ please show us your approach. Thanks in advance.
My attempt is think that I can compute for $n\geq 6$ with the known inequlity $$n(\log(n\log n)-1)<p_n<n\log(n\log n)$$ then using the first inequality $\log(n(\log(n\log n)-1))<\log(p_n)$ thus $$\frac{1}{\log p_n}<\frac{1}{\log(n(\log(n\log n)-1))},$$ also I need to use the second inequality, and binomial theorem to get the main term and the error term.
Arguing very naively, using $p_k \approx k \ln(k)$ and $\sum k^a \ln^b(k) \approx \int x^a \ln^b(x) dx $ ,
$\begin{array}\\ \sum_{k=1}^{n-1} p_{k} \sum_{j=k+1}^{n} p_{j} &\approx \sum_{k=1}^{n-1} k \ln(k) \sum_{j=k+1}^{n} j\ln(j)\\ &\approx \sum_{k=1}^{n-1} k \ln(k) \int_{x=k+1}^{n} x\ln(x)dx\\ &= \sum_{k=1}^{n-1} k \ln(k) ( \frac12 x^2\ln(x)-\frac14 x^2)\big|_{k+1}^{n} \\ &= \sum_{k=1}^{n-1} k \ln(k) ( \frac12 (n^2\ln(n)-k^2\ln(k))-\frac14 (n^2-k^2)) \\ &= (\frac12 n^2\ln(n)-\frac12 n^2) \sum_{k=1}^{n-1} k \ln(k) -\frac12\sum_{k=1}^{n-1} k^3\ln^2(k)+\frac14 \sum_{k=1}^{n-1} k^3 \ln(k) \\ &\approx (\frac12 n^2\ln(n)-\frac12 n^2) ( \frac12 n^2\ln(n)-\frac14 n^2) -\frac12( \frac14 n^4 \ln^2(x)-\frac18 n^2 \ln(x)+\frac1{32}n^4)) +\frac14 ( \frac14 n^4 \ln(n)-\frac1{16}n^4)) \\ &\approx \frac14 n^4\ln^2(n)-\frac38 n^4 \ln(n)+\frac18 n^4 - \frac18 n^4 \ln^2(x)+\frac1{16} n^2 \ln(x)+\frac1{64}n^4 + \frac1{16} n^4 \ln(n)-\frac1{64}n^4 \\ &= \frac18 n^4\ln^2(n)-\frac{5}{16} n^4 \ln(n)+\frac18 n^4 \\ &= \frac18 n^4(\ln^2(n)-\frac{5}{2} \ln(n)+1) \\ \end{array} $
I think there is a good chance that first term is correct, but I don't know about the others.