Which prime numbers is this inequality true for? Inequality
What values of $n$ satisfy the following inequality?

$$2(n-2) < p_n\prod_{i=3}^n \left(\frac{p_i-1}{p_i}\right)$$

Where $p$ are prime numbers and the notation $p_i$ indicates the $i$-th prime number.

See also, for some comments/answers for this
(https://mathoverflow.net/questions/234793/the-values-of-n-which-satisfy-an-inequality-about-prime-numbers/234840?noredirect=1#comment581295_234840)
 A: Take $n=10$ and $p_3=5,\ldots ,p_{10}={29}$. Then $2(n-2)=16$, but
$$
p_n\prod_{i=3}^n \frac{p_i-1}{p_i}=29\cdot \frac{5-1}{5}\cdot \frac{7-1}{7}\cdot \frac{11-1}{11}\cdot \frac{13-1}{13}\cdot \frac{17-1}{17}\cdot
\frac{19-1}{19}\cdot \frac{23-1}{23}\cdot \frac{29-1}{29}=13.741408409869844787061101504499000797,
$$
and $16<13.74\cdots $ is false. The infinite product (see here) is
$$
\displaystyle\prod_{i=1}^\infty\left(1-\dfrac{1}{p_i}\right)=\dfrac{1}{\zeta(1)},
$$
so that
$$
\prod_{i=3}^{\infty} \frac{p_i-1}{p_i}=0.
$$
I suppose that there are only a few $n$ where the inequality is true, i.e., for $n\le 5$. 
A: This is more of a long comment than an answer, but it might offer some insight beyond what's already in Dietrich Burde's answer.
Let's rewrite the OP's inequality with a slight shift in index and then some rearrangement of terms:
$$\begin{align}
2(n-1)\lt p_{n+1}\prod_{k=3}^{n+1}\left(p_k-1\over p_k\right)
&=p_{n+1}\left(4\over5\right)\left(6\over7\right)\left(10\over11\right)\cdots\left(p_{n+1}-1\over p_{n+1}\right)\\
&=4\left(6\over5\right)\left(10\over7\right)\cdots\left(p_{n+1}-1\over p_n\right)\\
&=3\left(1+{1\over3}\right)\left(1+{1\over5}\right)\left(1+{1\over7}\right)\cdots\left(1+{(p_{n+1}-p_n)-1\over p_n}\right)
\end{align}$$
This suggests an equivalent form for the OP's question:

What values of $n$ satisfy the inequality
$${2\over3}(n-1)\lt\prod_{k=1}^n\left(1+{g_k-1\over p_k}\right)$$
where $g_k=p_{k+1}-p_k$ is the gap to the next prime?

Note that including the prime $p_1=2$ has no effect, since $1+{g_1-1\over p_1}=1+{0\over2}=1$.  Note also that if the terms in the product were just $1+{g_k\over p_k}$, then the product would telescope down to $p_{n+1}\over2$, which is certainly greater than ${2\over3}(n-1)$.  Even if we just drop the $-1$ from the numerator $g_k-1$ for $k$ beyond some (fixed) value $N$, the inequality would reduce to something of the form ${2\over3}(n-1)\lt C_N{p_{n+1}\over p_N}$, which would eventually be satisfied, since $p_n\sim n\log n$.  So for the inequality to fail for most $n\gt5$, which is Dietrich's concluding supposition, the subtraction of $1$ from $g_k$ must continue to be significant, which suggests that a definitive answer must be based on some analysis of the prevalence of relatively small gaps between primes.
A: Write $x = p_n$, so that this inequality can be written in the form
\[2(\pi(x) - 2) < x \prod_{5 \leq p \leq x}\left(1 - \frac{1}{p}\right).\]
That is, the right-hand side is
\[3x \prod_{p \leq x} \left(1 - \frac{1}{p}\right).\]
By Mertens' third theorem, this is asymptotic to
\[\frac{3 e^{-\gamma_0} x}{\log x}\]
as $x \to \infty$, where $\gamma_0$ is the Euler-Mascheroni constant. In particular, $3 e^{-\gamma_0} \approx 1.68$.
On the other hand, the prime number theorem implies that the left-hand side is asymptotic to
\[\frac{2x}{\log x}\]
as $x \to \infty$. So the inequality is false for all sufficiently large $x$ (or equivalently, for all sufficiently large primes $p_n$). With a little extra effort, one can work out for which $x$ this inequality is first false, as in GH from MO's answer on MathOverflow.
