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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.

Ethan Bolker
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Chain Markov
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  • Comments, not a solution. First, $n=20$ is not a very large number of cases to base a conjecture on. Second, have you tried to work out what happens for the product of two primes? Then both summands on the left are easy to write down. – Ethan Bolker Aug 20 '18 at 15:02
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    The solution is contained in this answer: https://math.stackexchange.com/a/1215337/42969, where it is shown that $\sigma(n) + \phi(n) = 2 n + \sum_{d \vert n, d < n}{(a_d + 1) \cdot d}$, and that is $ \ge 2n$. – Martin R Aug 20 '18 at 15:03
  • As a remark, $\dfrac{\phi(n)+\sigma(n)}{n}$ can be arbitrarily large. This is because the series $$\sum_{p\text{ prime}},\frac{1}{p}$$ diverges, and $$\frac{\phi(n)+\sigma(n)}{n}>\sum_{i=1}^r,\frac{1}{p_i}$$ if the prime divisors of $n$ are $p_1,p_2,\ldots,p_r$. – Batominovski Aug 20 '18 at 18:25

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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, $\dfrac{\phi(n)}{n} + \dfrac{\sigma(n)}{n} \geq \prod_{i=1}^{r} (\dfrac{p_i-1}{p_i}) + \prod_{i=1}^{r} \bigg(\dfrac{p_i +1 }{p_i} \bigg) = \dfrac{\prod_{i=1}^{r}(p_i + 1) + \prod_{i=1}^{r} (p_i -1)}{\prod_{i=1}^{r} p_i} \geq \dfrac{2.\prod_{i=1}^{r} p_i }{\prod_{i=1}^{r} p_i} = 2$

subham saha
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  • I would like to add that the equality $\phi(n)+\sigma(n)=2n$ holds if and only if $n=1$ or $n$ is a prime natural number. – Batominovski Aug 20 '18 at 18:20