Edit [in response to a comment from anon]: Hereinafter, $N$ is a positive integer, $\sigma(N)$ is the sum-of-divisors of $N$, $\omega(N)$ is the number of distinct prime factors of $N$, and $\Omega(N)$ is the number of prime factors of $N$ (counting multiplicities).
Thus, $N$ is a perfect number if $\sigma(N) = 2N$.
A 2013 preprint by Nielsen claims to have proved that $\omega(N) \geq 10$ if $N$ is an odd perfect number. A paper by Ochem and Rao containing inequalities relating $\Omega(N)$ and $\omega(N)$ for $N$ an odd perfect number has been recently accepted in the Mathematics of Computation. The state-of-the-art result for $\Omega(N)$ remains to be Hare's $\Omega(N) \geq 75$.
[Edit - August 29] The state-of-the-art result for $\Omega(N)$ (where $N$ is an odd perfect number) is now Ochem and Rao's $\Omega(N) \geq 101$. [End edit]
[End edit - July 30 2013]
If there exists an $i \in \left[1,\omega(N)\right]$ such that
$$N \leq \frac{3}{2}{p_i}^{\alpha_i}\sigma({p_i}^{\alpha_i}),$$
then $$N = \prod_{i=1}^{\omega(N)}{{p_i}^{\alpha_i}}$$ (where the $p_i$'s are primes ordered in increasing magnitude and the $\alpha_i$'s are all positive)
is ${\it not}$ an odd perfect number. (See Theorem 4.2.5, page 112 in this M.Sc. thesis.)
In particular, suppose $i = 1$. (That is, let $p_1$ be the smallest prime factor of $N$.) Then we have
$${{p_1}^{\alpha_1}}\prod_{i=2}^{\omega(N)}{{p_i}^{\alpha_i}} = N = \prod_{i=1}^{\omega(N)}{{p_i}^{\alpha_i}} \leq \frac{3}{2}{p_i}^{\alpha_i}\sigma({p_i}^{\alpha_i}) < \frac{9}{4}{{p_1}^{2\alpha_1}},$$
from which it follows that
$$\prod_{i=2}^{\omega(N)}{{p_2}^{\alpha_i}} \leq \prod_{i=2}^{\omega(N)}{{p_i}^{\alpha_i}} < \frac{9}{4}{{p_1}^{\alpha_1}}.$$
But we also have
$${p_2}^{\Omega(N) - {\alpha_1}} = \prod_{i=2}^{\omega(N)}{{p_2}^{\alpha_i}} \leq \prod_{i=2}^{\omega(N)}{{p_i}^{\alpha_i}} < \frac{9}{4}{{p_1}^{\alpha_1}} < \frac{9}{4}{{p_2}^{\alpha_1}} < {{p_2}^{\alpha_1 + 1}},$$
from which we obtain
$$\frac{\Omega(N) - 1}{2} < \alpha_1.$$
Note that we have obtained the result:
"If $$N = {p_1}^{2\alpha_1}{q^k}\prod_{i=2}^{\omega(N) - 1}{{p_i}^{\alpha_i}}$$ is an odd (positive integer) with $\frac{\Omega(N) - 1}{2} < \alpha_1,$ then $N$ is ${\it not}$ perfect."
Taking the contrapositive of the result we have obtained, we have: "If $$N = {p_1}^{2\alpha_1}{q^k}\prod_{i=2}^{\omega(N) - 1}{{p_i}^{\alpha_i}}$$ is an odd perfect number with smallest prime factor $p_1$ and Euler prime $q$, then $\alpha_1 \leq \frac{\Omega(N) - 1}{2}$."
Somebody, please tell me that I ${\it did}$ make a logical error somewhere -- I am finding it increasingly hard to spot my own mistakes these days. =(
Thank you!