How to prove this inequality about primes? Let $p_n$ be the $n$-th prime. How to prove
$$2\prod_{i=2}^{n}\binom{p_i}{2}<\left(\frac{\prod_\limits{i=1}^{n}p_i}{\prod_\limits{i=1}^{n}p_i-\prod_\limits{i=2}^{n}(p_i-2)}\right)^{p_n^2}$$
for large n.

(this is not an answer yet)
Taking the logarithm :
$$\ln 2\prod_{i=1}^n {p_i \choose 2}= -(n-1)\ln 2+\sum_{i=1}^n \ln p_i+\ln (p_i-1)$$
$$=(1-n)\ln 2+2\sum_{i=1}^n \ln p_i+\sum_{i=1}^n \ln(1-\frac{1}{p_i})$$
$$\sim
(1-n)\ln 2+2\sum_{i=1}^n \ln p_i-\sum_{i=1}^n \frac{1}{p_i}\sim (1-n)\ln 2+ 2p_n- \ln \ln p_n$$
where I used that $\sum_{p < x} \ln p \sim x$ and $\sum_{p < x} \frac{1}{p} \sim \ln \ln x$ that is the prime number theorem and the Mertens theorem
what do you get following the same lines for the RHS ?
