Suppose $\phi$ is Euler totient function and $\sigma$ is divisor sum. Is $\phi(n) + \sigma(n) \geq 2n$ true for every natural $n$?
I manually checked the inequality for all numbers between $1$ and $20$ - and it holds on them. I do not know, however, how to prove this fact in general.
Also, it is not hard to see, that for prime $p$, $\phi(n) + \sigma(n) = 2n$. That means, that if a counterexample exists, it has to be composite.
Any help will be appreciated.
Let $n = p_1^{\alpha_1}\cdots p_r^{\alpha_r}$.
$\dfrac{\phi(n)}{n} = \prod_{i=1}^{r} (\dfrac{p_i-1}{p_i})$
$\dfrac{\sigma(n)}{n} = \prod_{i=1}^{r} \Bigg(\dfrac{p_i-\dfrac{1}{p_i^{\alpha_i}}}{p_i-1} \Bigg) \geq \prod_{i=1}^{r} \Bigg(\dfrac{p_i-\dfrac{1}{p_i}}{p_i-1} \Bigg) = \prod_{i=1}^{r} \bigg(\dfrac{p_i +1 }{p_i} \bigg)$
So, $\dfrac{\phi(n)}{n} + \dfrac{\sigma(n)}{n} \geq \prod_{i=1}^{r} (\dfrac{p_i-1}{p_i}) + \prod_{i=1}^{r} \bigg(\dfrac{p_i +1 }{p_i} \bigg) = \dfrac{\prod_{i=1}^{r}(p_i + 1) + \prod_{i=1}^{r} (p_i -1)}{\prod_{i=1}^{r} p_i} \geq \dfrac{2.\prod_{i=1}^{r} p_i }{\prod_{i=1}^{r} p_i} = 2$