Does this inequality hold true, in general? Let
$$N = \prod_{i=1}^{\omega(N)}{{p_i}^{\alpha_i}}$$
be the prime factorization of the positive integer $N$.
Does the following inequality hold true in general?
$$\prod_{i=1}^{\omega(N)}{\frac{p_i}{p_i - 1}} < \frac{2\prod_{i=1}^{\omega(N)}{p_i}}{1 + \prod_{i=1}^{\omega(N)}{p_i}}$$
If not, for what particular values of $N$ does this inequality hold?
Note that $\omega(N)$ denotes the number of distinct prime factors of $N$.
 A: We can cancel the
numerator of each side to get
the equivalent inequality
$\prod_{i=1}^{\omega(N)}{\dfrac{1}{p_i - 1}} < \dfrac{2}{1 + \prod_{i=1}^{\omega(N)}{p_i}}
$
or
$1 + \prod_{i=1}^{\omega(N)}{p_i}
< 2\prod_{i=1}^{\omega(N)}(p_i - 1) 
$.
Let's look at
the sum over primes
$f(x) =
\sum_{p < x} \ln (p-1)
$.
Since
$\sum_{p < x} \ln p \approx  x$
(from the prime number theorem),
$\begin{align}
f(x) 
&=\sum_{p < x} \ln (p-1)\\
&=\sum_{p < x} (\ln p+\ln(1-1/p))\\
&=\sum_{p < x} \ln p+\sum_{p < x}\ln(1-1/p)\\
&\approx x-\sum_{p < x}(1/p+1/(2p^2)+...)\\
&\approx x-\ln \ln x+C\\
\end{align}
$
From this,
$e^{f(x)}
\approx e^{x-\ln \ln x+C}
= C_1 e^x/\ln x
$.
If 
$N(x) = \prod_{p < x} p$
and
$M(x) = \prod_{p < x} (p-1)$,
$N(x) \approx e^x$
and
$M(x) \approx C_1 e^x/\ln x$,
$1+N(x) < 2M(x)$
means,
for large enough $x$,
$e^x < C_1 e^x/\ln x$,
which is false.
So that inequality is often false.
A: It turns out that
that inequality is true
about 95% of the time.
As in my previous answer,
 cancel the
numerator of each side to get
the equivalent inequality
$\prod_{i=1}^{\omega(N)}{\dfrac{1}{p_i - 1}} < \dfrac{2}{1 + \prod_{i=1}^{\omega(N)}{p_i}}
$
or
$1 + \prod_{i=1}^{\omega(N)}{p_i}
< 2\prod_{i=1}^{\omega(N)}(p_i - 1) 
$.
Dividing by 
$\prod_{i=1}^{\omega(N)}{p_i}$,
and, using the fact that
$\dfrac{\phi(n)}{n}
=\prod_{i=1}^{\omega(N)}\dfrac{p_i-1}{p_i}
$,
this becomes
$1+\dfrac{1}{\phi(n)}
<2\dfrac{\phi(n)}{n}
$.
Since $\dfrac{1}{\phi(n)}$
is small for large $n$,
the proportion of $n$ satisfying this
is the same as the proportion satisfying
$\frac{1}{2}
<\dfrac{\phi(n)}{n}
$,
or
$2
>\dfrac{n}{\phi(n)}
$.
At this point,
I did a Google search for
"density of euler phi function".
The second link is
http://www.ams.org/journals/proc/2007-135-09/S0002-9939-07-08771-0/S0002-9939-07-08771-0.pdf.
This paper,
by ANDREAS WEINGARTNER,
is titled
"THE DISTRIBUTION FUNCTIONS OF σ(n)/n AND n/ϕ(n)".
Here is its abstract:
"Let σ(n) be the sum of the positive divisors of n. We show that
the natural density of the set of integers n satisfying σ(n)/n ≥ t is given
by 
$\exp\big(−e^{t e^{−γ}(1 + O(t^{−2}))}\big)$
, where γ denotes Euler’s constant. The same
result holds when σ(n)/n is replaced by n/ϕ(n), where ϕ is Euler’s totient
function."
This paper has clearly done
the heavy lifting.
If we put $t = 2$,
and use $\gamma \approx 0.5772156649$
(I show each stage in the computation for checkability),
$te^{-\gamma} = 1.1229189671$,
$e^{te^{-\gamma}}=3.0738134815$,
$\exp(-e^{te^{-\gamma}})
=0.0462444658
$.
This is the density of $n$ for which
$\dfrac{n}{\phi(n)}
> 2
$.
The density for which
$\dfrac{n}{\phi(n)}
< 2
$
is one minus this
or $0.9537555342$.
