# Lower bound of Euler phi function times sum of divisors

After some work, I got this nice inequality:

$$\frac{n^2}{2} < \phi(n)\cdot \sigma(n)$$

where $\phi(n)$ is Euler's phi function and $\sigma(n)= \sum_{d|n} d$. I know this is true because I'm aware that this can be further refined to

$$\frac{6 n^2}{\pi^2} < \phi(n)\cdot \sigma(n)$$

However, I'm interested in the first one because I'm sure there is an elemental proof of it (which I can't find at the moment). Any ideas?

• Note if you want the $\varphi$ instead of $\phi$ It's written as \varphi – kingW3 Dec 10 '14 at 20:03
• I've seen it written both ways. Is there any standard way of writing it? – nabla Dec 10 '14 at 20:06

If $n=\prod_ip_i^{a_i}$, then $$\sigma(n)=\prod_i \frac{p_i^{a_i+1}-1}{p_i-1}=n\prod_i\frac{1-p_i^{-a_i-1}}{1-p_i^{-1}},$$ and $$\phi(n)=n\prod_i(1-p_i^{-1})$$ Hence we obtain $$\frac{\sigma(n)\phi(n)}{n^2}=\prod_i (1-p_i^{-a_i-1}).$$ Hence the first inequality is obvious, and the second also: each of the exponents is less than or equal to $−2$, so the product is at least as large as the product 􏰌$\prod_p(1 − p^{−2})=\frac{1}{\zeta(2)}=\frac{6}{\pi^2}$. Hence we obtain $$6\frac{n^2}{\pi^2}<\sigma(n)\phi(n).$$ The first inequality is obtained if we just use $\frac{1}{\zeta(2)}>\frac{1}{2}$.