Are primes less than the sum of divisors?

Question

I am trying to prove that

Let $p_n$ be the $n$th prime number, $\sigma (n)=\sum_{d|n}d$. Prove that $$\sigma(n) \le p_n$$

It seems obvious at first glance-to me, at least the sum of divisors of $n$ is less than the $n$th prime. In fact, it was simple to prove for numbers that are deficient or perfect, but it seems difficult to prove when $\sigma (n)>2n$.

One thing I discovered was that $$\frac{\sigma (n)}{n}=\sum_{d|n}\frac{d}{n}=\sum_{d|n}\frac{1}{d} \le H_n \Leftrightarrow \sigma(n) \le nH_n$$ Where $H_n$ is the harmonic number. I am unsure how to proceed, could anyone provide assistance? Thanks.

Answer 1

G. Robin showed (1984) unconditionally that, for $n \geq 3,$ $$ \sigma(n) \leq e^\gamma n \log \log n + \frac{0.6482 \; \; n}{\log \log n} $$ with the constant $0.6482...$ chosen to give equality for $n=12.$

Rosser and Schoenfeld (1962) showed that, for $n \geq 2,$ $$ p_n > n \left( \log n + \log \log n - \frac{3}{2} \right) $$