Let
$$N = \prod_{i=1}^{w}{{p_i}^{\alpha_i}}$$
be the prime factorization of the positive integer $N > 1$, where $w = \omega(N)$ is the number of distinct prime factors of $N$.
Let $\sigma(N)$ be the sum of divisors of $N$, and let $\varphi(N)$ be the Euler totient of $N$.
Then we have
$$\dfrac{\sigma(N)}{N}=\dfrac{\displaystyle\prod_{i=1}^{w}{\sigma\bigg({p_i}^{\alpha_i}\bigg)}}{\displaystyle\prod_{i=1}^{w}{{p_i}^{\alpha_i}}} = \displaystyle\prod_{i=1}^{w}{\dfrac{{p_i}^{\alpha_i + 1} – 1}{{p_i}^{\alpha_i}(p_i – 1)}} < \displaystyle\prod_{i=1}^{w}{\dfrac{{p_i}^{\alpha_i + 1}}{{p_i}^{\alpha_i}(p_i – 1)}} = \displaystyle\prod_{i=1}^{w}{\dfrac{p_i}{p_i – 1}} = \displaystyle\prod_{i=1}^{w}{\dfrac{{p_i}^{\alpha_i}}{{p_i}^{\alpha_i – 1}(p_i – 1)}} = \dfrac{\displaystyle\prod_{i=1}^{w}{{p_i}^{\alpha_i}}}{\displaystyle\prod_{i=1}^{w}{\varphi\bigg({p_i}^{\alpha_i}\bigg)}} = \dfrac{N}{\varphi(N)}.$$
Here is my question:
Can we improve on the inequality $\sigma(N)\varphi(N) < N^2$ for integers $N > 1$?
Best Answer
We have pretty many such $N$ that $\sigma(N)\varphi(N)=N^2−1$. They are known as prime numbers. So as you can see, there is actually no room for improvement.