# Lower bound for the sum of divisor function

Let $\sigma$ be the sum-of-divisors function defined by $$\sigma(n) = \sum_{d \mid n} d.$$ Is there an explicit lower bound for $\sigma(n)/n$ in the style of the lower bound $$\phi(n) > \dfrac{n}{e^\gamma \log \log n + \frac{3}{\log \log n}}$$ for $n > 2$, where $\phi$ is the Euler's totient function and $\gamma$ is the Euler-Mascheroni constant? I'm aware that the inequality $$\dfrac{6}{\pi^2} < \dfrac{\phi(n) \sigma(n)}{n^2} < 1,$$ for $n > 1$, is known. But can we get a better lower bound than $\sigma(n)/n > 6n/\pi^2\phi(n)$?

• Interesting upper bounds, yes. Lower bounds, no. Whenever $n$ is prime, $$\frac{\sigma(n)}{n} = 1 + \frac{1}{n}.$$ – Will Jagy May 28 '14 at 0:57
• upper bounds, short version at math.stackexchange.com/questions/808168/… – Will Jagy May 28 '14 at 1:03

You can't do better if you want a lower bound that holds for all $n$. If you consider the sequence of integers which are products of the consecutive primes, i.e. $$2, \;2\cdot3, \;2\cdot3\cdot5, \;2\cdot3\cdot5\cdot7,\,\dots$$ you can see that your bound is sharp.