Show that $\displaystyle\frac{n}{\phi(n)}\geq\sum_{m\leq k}\frac{1}{m}$ Suppose that for all primes $p\leq k$ we have  $p|n$ where $n\in\mathbb N$
How can I show that $\displaystyle\frac{n}{\phi(n)}\geq\sum_{m\leq k}\frac{1}{m}$ where $\phi(n)$ is the Euler Totient function.
$$\frac{n}{\phi(n)}=\frac{1}{\prod_{p|n} (1-\frac{1}{p})}=\prod_{p|n}\frac{1}{1-\frac{1}{p}}=\prod_{p|n}\left(1+p^{-1}+{p^{-2}}+\dots\right)$$
But not sure where to go from here or if this is even the right way to start. Any hints?

Let $p_1,p_2,\dots,p_r$ be the primes $\leq k$ then any positive integer $m\leq k$ has a unique prime factorization 
$$m=p_1^{j_1}\cdot p_2^{j_2}\cdots p_r^{j_r}.$$
Hence $1/m$ can be obtained in a unique way by multiplying just the term $1/p_i^{j_i}$ of the $i$-th factor of product 
$$\prod_{i=1}^r\left(\sum_{j=0}^{\infty}\frac{1}{p_i^{j}}\right)$$
for $i=1,2,\dots,r$.
Finally, by your work, and since $p_i| n$ for $i=1,2,\dots,r$, we find that
$$\frac{n}{\phi(n)}=\prod_{p|n}\left(\sum_{j=0}^{\infty}\frac{1}{p^{j}}\right)\geq 
\prod_{i=1}^r\left(\sum_{j=0}^{\infty}\frac{1}{p_i^{j}}\right)\geq\sum_{m\leq k}\frac{1}{m}.$$
