Show the inequality involving the product of distinct primes that divide $n$. Let $\displaystyle \omega (n)=\sum_{p\mid n} 1$ be the number of distinct primes dividing $n$. Show that
$\displaystyle \phi (n) \geq n \prod_{k=2}^{\omega (n) +1} (1 - \frac{1}{k}) = \frac{n}{\omega (n) +1}$ and that 
$\displaystyle \phi (n) > \frac{cn}{\text{log } n}$ (hint: show that $2^{\omega (n)} \leq \tau (n) \leq n$.)
This is a problem on a practice sheet for a section in Leveque's Number Theory. I had another question from this sheet that said to use induction, but it turned out that the solution did not need induction. Is this problem manageable with just knowing the following? If so, I am not sure on where to start and proceed. 
$\displaystyle \tau (n) = \sum_{d|n} 1$.
$\tau (p^e) = e+1$. 
$\tau$ is multiplicative.
$\displaystyle \sigma (n) = \sum_{d|n} d$.
$\displaystyle \sigma (p^e) = 1+p+p^2+...+p^e = \frac{p^{e+1}-1}{p-1}$. 
$\sigma$ is multiplicative.

First, suppose that $n = \prod\limits_{k = 1}^{ω(n)} p_k^{a_k}$, then\begin{align*}
\frac{φ(n)}{n} &= \prod_{k = 1}^{ω(n)} \left( 1 - \frac{1}{p_k} \right) \geqslant \prod_{k = 1}^{ω(n)} \left( 1 - \frac{1}{k + 1} \right) = \prod_{k = 2}^{ω(n) + 1} \left( 1 - \frac{1}{k} \right)\\
&= \prod_{k = 2}^{ω(n) + 1} \frac{k - 1}{k} = \frac{1}{ω(n) + 1},
\end{align*}
i.e.$$
φ(n) \geqslant n \prod_{k = 2}^{ω(n) + 1} \left( 1 - \frac{1}{k} \right) = \frac{n}{ω(n) + 1}.
$$
Now, because the positive divisors of $n$ are positive integers no greater than $n$, then $τ(n) \leqslant n$. Also,$$
τ(n) = \prod_{k = 1}^{ω(n)} (1 + a_k) \geqslant \prod_{k = 1}^{ω(n)} 2 = 2^{ω(n)}.
$$
Therefore,$$
φ(n) \geqslant \frac{n}{ω(n) + 1} \geqslant \frac{n}{\log_2 τ(n) + 1} \geqslant \frac{n}{\log_2 n + 1} > \frac{cn}{\ln n}.
$$
