Proving $\sigma_1(n) \phi(n)>n/2$

Question

I need to prove that $\sigma_1(n) \phi(n)$ is less or equal than $n^2$ and greater than $n/2$. I managed to prove the first part but now I need some help on proving that it's greater than $n/2$. I got to $\sigma_1(n) \phi(n)=\prod_{i=1}^s p_i^{2 \alpha_i}-p_i^{\alpha_i-1}$.

$n=\prod_{i=1}^s p_i^{\alpha_i}$.

Answer 1

What you have is enough to prove the second part, consider $$2\frac{\sigma_1 (n) \phi (n)}{n} = 2 \prod_{i = 1}^s \frac{p_i^{\alpha_i +1} - 1}{p_i}$$ Since $\alpha_i \geqslant 1$, you have $$2\frac{\sigma_1 (n) \phi (n)}{n} \geqslant 2 \prod_{i = 1}^s \frac{p_i^2 - 1}{p_i} = 2 \prod_{i = 1}^s \frac{(p_i - 1)(p_i+1)}{p_i} \geqslant 2$$

Maybe am I mistaken?

Answer 2

From $\sigma_1(n)\geq n$ and $\phi(n)\geq1$ we have $\sigma_1(n)\phi(n)\geq n>\frac{n}{2}$.

Answer 3

The estimate follows directly from $\sigma(n)\ge n$. Actually, we have $$ \frac {6}{\pi^2} < \frac{ \sigma(n)\phi(n)}{n^2} < 1 $$ for all $n>1$. Did you really mean $\sigma(n)\phi(n)>\frac{n}{2}$? It looks like a typo, because we immediately have $\sigma(n)\phi(n)\ge n$, so why bother with $\frac{n}{2}$?