# On the relationship between $\phi(n)$ and $\sigma( n)$

I recently learnt that $\frac{\sigma(n)}{n} \leq \frac{n}{\phi(n)}$, were $\sigma(n)$ denotes the divisor function, $\phi(n)$ the Euler totient function and $n\geq 2$ is an integer.

My questions is : When does equality hold, and is there an integer $n_0$ such that $\frac{\sigma(n_0)}{n_0} < \frac{n_0}{\phi(n_0)}$ for all $n\geq n_0$ ?

A proof or reference will be most welcome.

Note that both sides are multiplicative, so you get equality only if you get equality for the prime powers that divide $n$.
But \begin{align}\sigma(p^k)\phi(p^k) &= \frac{p^{k+1}-1}{p-1}p^{k-1}(p-1)\\ &=(p^{k+1}-1)(p^{k-1}-1)\\ &=p^{2k}-p^{k+1}-p^{k-1}+1\\ &\leq p^{2k}-p^{k+1}<(p^k)^2 \end{align} So no equality.
Wikipedia says that $$\frac {6}{\pi^2} < \frac{ \sigma(n)\phi(n)}{n^2} < 1$$ for all $n>1$.
So, equality in $\dfrac{\sigma(n)}{n} \leq \dfrac{n}{\phi(n)}$ holds only for $n=1$ and strict inequality for all $n\ge 2$.