How can I prove this inequality involving the primes? I want to determine whether
$${{3({p_n}-p_{n-1})}\over{p_{n-1}}}\ge\prod_{i=3}^{n-1}\Bigg(1-{2\over{p_i}}\Bigg)$$
is true for all sufficiently large $n\gt3$. (I don't know whether or not it's actually true, but it tests as true for a bunch of small $n$ that I've looked at).
My first thought was to use the lower-bound approximation for the $n$'th prime $n[ln(n)+ln(ln(n))-1]$ as $p_n$ on the left side, and the upper-bound approximation $n[ln(n)+ln(ln(n))]$ as $p_n$ on the right side. That at least gets rid of the primes, but of course the product on the right side still isn't smooth, so I don't know where to go from there. Also, I'm not sure if the inequality would remain true with those approximations, if it's indeed true with the primes. (I did test a bunch of small values and it seemed to remain true, but again I know that means nothing.)
Thanks for any help.
 A: It's not true.
I happen to have a file of the first 1400 primes sitting around so I created a spreadsheet for the LHS and the RHS.  The LHS is less than the RHS for twin primes 43 and higher.
Note I started the product on the RHS at $p_n=5$ since it indicates $i=3$ not $p=3$.  If you start the product with $p=3$ it takes until 229 for the first violation.
Note that merely reversing the inequality will not create a viable statement as written, since the gaps between primes vary.  Thus although the twin primes violate the original statement at the number given, there are many others that continue to meet the condition given.
A: This is just an elaboration of my comment above, definitely not a solution.
It is fairly easy to show that $$\prod_{i=3}^{n}\frac{1-2/p_i}{(1-1/p_i)^2}=\prod_{i=3}^n\left(1-\frac{1}{(p_i-1)^2}\right)$$ is decreasing and has a non-zero limit as $n\to\infty$. 
Let $\alpha$ be that limit. It looks like $\alpha \approx 0.880$.
For any $\epsilon>0$, then, for large enough $n$, $$9\alpha a_n^2<\prod_{i=3}^n (1-2/p_i) <9(\alpha+\epsilon)a_n^2$$
This might make it easier, since $\{a_i\}$ is a fairly well-known sequence.
where $a_n=\prod_{i=1}^n (1-1/p_i)$. (The factor of $9$ comes because we've added the cases of $i=1,2$.)
So, at minimum, you need:
$$a_n^2 \leq \frac{2}{3\alpha}\frac{p_{n+1}-p_n}{p_n}$$
for sufficiently large $n$.
A: Estimate of the Product
Using the Prime Number Theorem, we get
$$
\begin{align}
\sum_{k=1}^n\frac1{p_k}
&=\int_1^{p_n}\frac1x\,\mathrm{d}\pi(x)\\
&=\frac{\pi(p_n)}{p_n}+\int_2^{p_n}\frac{\pi(x)}{x^2}\,\mathrm{d}x\\
&=\frac{n}{p_n}+\int_2^{p_n}\frac{\frac x{\log(x)}+O\left(\frac x{\log(x)^2}\right)}{x^2}\,\mathrm{d}x\\[18pt]
&=\log(\log(p_n))+O(1)\tag{1}
\end{align}
$$
Applying $(1)$, we have
$$
\begin{align}
\log\left(\prod_{i=3}^{n-1}\left(1-\frac2p_i\right)\right)
&=\sum_{i=3}^{n-1}\log\left(1-\frac2p_i\right)\\
&=\sum_{i=3}^{n-1}\left(-\frac2p_i+O\left(\frac1{p_i^2}\right)\right)\\[12pt]
&=-2\log(\log(p_{n-1}))+O(1)\tag{2}
\end{align}
$$
Therefore,
$$
\prod_{i=3}^{n-1}\left(1-\frac2p_i\right)=\Theta\left(\frac1{\log(p_{n-1})^2}\right)\tag{3}
$$
Estimate of the Ratio
We can derive from the Prime Number Theorem that $p_n=n\log(n)+O(n\log(\log(n)))$. This shows that for any $\alpha\gt0$
$$
\lim_{n\to\infty}\frac{p_{\alpha n}}{p_n}=\alpha\tag{4}
$$
Furthermore, $(4)$ shows that $\lim\limits_{n\to\infty}\frac{p_n}{p_{n-1}}=1$, therefore, $\frac{p_k-p_{k-1}}{p_{k-1}}\sim\log\left(\frac{p_k}{p_{k-1}}\right)$ Thus, for any $\epsilon>0$, there is an $N$ so that for $m\ge N$,
$$
\begin{align}
m
&=\pi(p_m)\\
&\ge(1-\epsilon)\frac{p_m}{\log(p_m)}\tag{5}
\end{align}
$$
and
$$
\begin{align}
\sum_{k=m+1}^n\frac{p_k-p_{k-1}}{p_{k-1}}
&\le(1+\epsilon)\sum_{k=m+1}^n\log\left(\frac{p_k}{p_{k-1}}\right)\\
&=(1+\epsilon)\log\left(\frac{p_n}{p_m}\right)\tag{6}
\end{align}
$$
Thus, for any $n\ge2N$, there must be a $\frac n2\lt k\le n$ so that
$$
\begin{align}
\frac{p_k-p_{k-1}}{p_{k-1}}
&\le(1+\epsilon)\frac{2\log(2)}{n}\\
&\le(1+\epsilon)\frac{2\log(2)}{k-1}\\
&\le\frac{1+\epsilon}{1-\epsilon}2\log(2)\frac{\log(p_{k-1})}{p_{k-1}}\tag{7}
\end{align}
$$
Comparing $(3)$ and $(7)$, there is no constant of proportionality that would make the given inequality true for all $n$.
A: First, I replaced $p_n - p_{n-1}$ with $2$ since there are infinitely many twin primes.
Then, by incrementing $n$ and rearranging, I get $$\frac6{p_{n-1}} \cdot \frac{p_{n-1}}{p_n} \ge \left( 1-\frac2{p_n} \right) \cdot \prod_{i=3}^{n-1} \left( 1-\frac2{p_i} \right)$$ 
Hmm, just saw a problem with the next step - I'll come back and edit after working on it.
