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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.

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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.

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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 $n\ge 2$.

Wikipedia cites the book by Hardy & Wright, theorem 329. The relevant page in the fourth edition is reproduced below.


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