Upper Bound for Finite Euler Product Let $n$ be a positive integer. I am trying to show
$$\prod_{p\le n/2, p\,\text{prime}}(1-1/p)\le \frac{1+\pi(n)}{n},$$
where $\pi(n)$ denotes the number of primes $\le n$.
After doing some calcuations by hand for small $n$, this seems clear. According to some Mathematica computations, this is true for $n\le 20,000$. However, I am not sure of how to give a rigorous proof since I am not aware of any upper bounds for the Euler product. I know of asymptotics for the Euler product, but I am not interested in what happens in the limit since I'd like to see if this is true for each $n$.
One idea is to apply the fundamental theorem of arithmetic, to write $$\prod_{p\le n/2, p\,\text{prime}}(1-1/p)=\frac{1}{\sum_{i\in F_{n/2}}\frac{1}{i}},$$
where $F_{n/2}$ is the collection of positive integers with greatest prime factor at most $n/2$, but I am unsure of the size of this sum compared to $\frac{1+\pi(n)}{n}.$
 A: Since $1-x\leq e^{-x}$ for any $x\in\left[0,\frac{1}{2}\right]$, we clearly have
$$ \prod_{\substack{p\text{ prime}\\p\leq n/2}}\left(1-\frac{1}{p}\right)\leq \exp\Big[-\!\!\!\sum_{\substack{p\text{ prime}\\p\leq n/2}}\frac{1}{p}\Big] $$
and the magnitude of the RHS is $\frac{1}{\log(n/2)}$ by Mertens' second Theorem. This is also the magnitude of $\frac{\pi(n)}{n}$ by the PNT, but I honestly do not know if the original inequality can be proved directly.
A: By Dusart proof we know that 
$\prod \limits_{p \leq \frac{n}{2}} 1-\frac{1}{p} \leq \frac{e^{-\gamma}}{\ln n-\ln 2} (1+\frac{0.2}{(\ln n-\ln 2)^2})$ for all $n \geq 5946$ and $\gamma \approx 0.577$ its Euler constant.
On the other hand we have that $\frac{1+\pi(n)}{n} \geq \frac{\pi(n)}{n} \geq \frac{\frac{n}{\ln n}}{n} $ for all $n \geq 17$ by J.Rosser result.
So for all $n \geq 5946$ we need to prove that $\frac{e^{-\gamma}}{\ln n-\ln 2} (1+\frac{0.2}{(\ln n-\ln 2)^2}) < \frac{\frac{n}{\ln n}}{n} = \frac{1}{\ln n}$
Its easy to see that $\frac{0.2}{(\ln n-\ln 2)^2} \leq \frac{0.2}{(\ln 5946-\ln 2)^2} \leq 0.004$ for all $n \geq 5946$ ($\ln n$ is monotonic increasing).
So we need to prove that $\frac{e^{-\gamma}}{\ln n-\ln 2}(1+0.004) < \frac{1}{\ln n}$, since $e^{-\gamma} *(1+0.004) < 0.6$ we get that $\frac{0.6}{\ln n-\ln 2} < \frac{1}{\ln n}$ which is just $0.6 \ln n < \ln n-\ln 2$ for all $n \geq 3$.
Meaning that $\ln n^{0.6} < \ln 0.5n$ exponent-ate both sides to get that $n^{0.6} \leq 0.5n$ which is true for all $n\geq 6$,
So the inequality holds true for all $n \geq 5946$, and checking for smaller integers the inequality holds true for all integers $n>3$
Note : that since $\frac{0.2}{(\ln n-\ln 2)^2}$ vanishes then the ratio between $\frac{1+\pi(n)}{n}$ over $\prod \limits_{p \leq \frac{n}{2}} 1-\frac{1}{p} $ is equal to $e^{\gamma} \approx 1.78107$ as $n \to \infty$
Done.
