TITLE: How can I prove this inequality involving the primes?
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: 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 $n$.