PROBLEM STATEMENT (ORIGINAL)
If $p$ is prime, $m$ is composite and deficient, and $I(p) < I(m)$, does it follow that $m < p$, where $I(x)=\sigma(x)/x$ for $x \in \mathbb{N}$?
MY ATTEMPT
Let $$m = \prod_{i=1}^{r}{{q_i}^{\alpha_i}},$$ where $q_1 < \ldots < q_m$ are primes, for some $r = \omega(m) \geq 2$.
Then $$I(m) = \prod_{i=1}^{r}{I({q_i}^{\alpha_i})}$$ and we are given that $$\frac{p+1}{p} = I(p) < I(m).$$ It follows that $$p > \frac{1}{I(m) - 1} = \frac{1}{\prod_{i=1}^{r}{I({q_i}^{\alpha_i})} - 1} = \bigg(\prod_{i=1}^{r}{I({q_i}^{\alpha_i})} - 1\bigg)^{-1}.$$
I believe that the Bernoulli Inequality does not apply in this case. This is where I get stuck.
PROBLEM STATEMENT (REVISED)
If $p$ is prime, $m$ is composite and deficient, and $I(p) < I(m)$ with $p \nmid m$, does it follow that $m < p$, where $I(x)=\sigma(x)/x$ for $x \in \mathbb{N}$?