Does a positive constant $\nu$ exist so that $\varphi(n)>\nu\cdot n$ for all $n$? Does a positive constant $\nu$ exist so that $\varphi(n)>\nu\cdot n$ for all $n$? Clearly this problem is exactly the same as asking if $\prod\limits_{i=1}^\infty \frac{p_i-1}{p_i}=0$. This is because $\varphi(n)=n\prod\limits_{p|n}\frac{p-1}{p}$. And if $n$ has $k$ prime divisors this is clearly greater than $n\prod_{i=1}^k\frac{p_i-1}{p_i}$.
So the problem breaks down to figuring out to what the sequence $a_n=\prod\limits_{i=1}^n\frac{p_i-1}{p_i}$ converges. (It clearly converges since it is decreasing and bounded). In fact we only need to figure out if it converges to zero or something else.
 A: If $\prod_{k=1}^n\left(1-\frac{1}{p_i}\right)$ converges to a non-zero value, then:
$$\prod_{k=1}^n \frac{1}{1-\frac{1}{p_i}}$$ converges. Then write $$\frac{1}{1-\frac{1}{p_i}}=\sum_{j=0}^\infty \frac{1}{p_i^j}$$
Then $$\prod_{k=1}^n\frac{1}{1-\frac{1}{p_i}} = \sum_{m} \frac{1}{m}$$
where the sum is over the natural numbers $m$ that have no prime factors other than $p_1,\dots,p_n$.
In particular, then, $$\prod_{k=1}^n\frac{1}{1-\frac{1}{p_i}} > \sum_{m=1}^{n} \frac{1}{m}$$
And $\lim_{n\to\infty} \sum_{m=1}^{n}\frac{1}{m}=+\infty$.
A: Rosser and Schoenfeld (1962) prove that, for $n \geq 3,$
$$ \frac{n}{\phi(n)} \leq e^\gamma \log \log n  + \frac{2.50637}{\log \log n}, $$
where the constant $2.50637$ is the value that gives equality at the primorial
$$ 223092870  $$
and nowhere else.
Meanwhile, Nicolas (1983) that the condition that 
$$ \frac{N}{\phi(N)} > e^\gamma \log \log N   $$
for all primorials $N$ is equivalent to the Riemann Hypothesis.
Here we go, Hardy and Wright, theorem 328, page 267 in my edition, result of Landau that
$$  \liminf \frac{\phi(n) \log \log n}{n} = e^{-\gamma}  $$
That is, there is a sequence of numbers with
$$ \phi(n) \approx n \left( \frac{e^{-\gamma} }{  \log \log n} \right) $$
and no such constant $\nu$ exists.
A: Since $\displaystyle \sum_i \dfrac{1}{p_i} = \infty$, $\displaystyle \prod_i \left(1 - \dfrac{1}{p_i}\right) = 0$.
