Let $x \in \mathbb{N}$, and let $$x = \prod_{i=1}^{\omega(x)}{{p_i}^{\alpha_i}}$$ be the canonical factorization of $x$. (That is, the $p_i$'s are primes with $p_1 < \ldots < p_{\omega(x)}$.) Note that $\omega(x)$ is the number of distinct prime factors of $x$.
What is wrong with this proof that $p_1 > \omega(x)$, where $p_1$ is the least prime dividing $x$?
Let $\sigma(x)$ be the sum of the divisors of $x$. Then the abundancy index $I(x) = \sigma(x)/x$ is bounded above by $$I(x) = I\left(\prod_{i=1}^{\omega(x)}{{p_i}^{\alpha_i}}\right) = \prod_{i=1}^{\omega(x)}{I\left({p_i}^{\alpha_i}\right)} < \prod_{i=1}^{\omega(x)}{\left(\frac{p_i}{p_i - 1}\right)}.$$
Now, I use the inequality $$\frac{p_i}{p_i - 1} < \frac{p_1 - (i - 1)}{p_1 - i}$$ for $i \neq 1$. This yields $$1 \leq I(x) < \prod_{i=1}^{\omega(x)}{\left(\frac{p_i}{p_i - 1}\right)} = \left(\frac{p_1}{p_1 - 1}\right)\prod_{i=2}^{\omega(x)}{\left(\frac{p_i}{p_i - 1}\right)} < \left(\frac{p_1}{p_1 - 1}\right)\prod_{i=2}^{\omega(x)}{\left(\frac{p_1 - (i - 1)}{p_1 - i}\right)} = \frac{p_1}{p_1 - \omega(x)}.$$
Finally, I have $p_1 > \omega(x)$.