A question about odd perfect numbers 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!
 A: You've written
(1) 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 $N$ (where the $p_i$'s are primes ordered in increasing magnitude and the $\alpha_i$'s are all positive) is not an odd perfect number.
(2) If $N \leq \frac{3}{2}{p_1}^{\alpha_1}\sigma({p_1}^{\alpha_1})$ where $p_1$ is the smallest prime factor of $N$, then $\frac{\Omega(N) - 1}{2} < \alpha_1$.
(3) We have obtained the result: "If $N$ is an odd positive integer with $\frac{\Omega(N) - 1}{2} < \alpha_1,$ then $N$ is not perfect."
(4) Taking the contrapositive of the result we have obtained : "If $N$ is an odd perfect number with smallest prime factor $p_1$ and Euler prime $q$, then $\alpha_1 \leq \frac{\Omega(N) - 1}{2}$."
Now, let 
$\qquad A$ : $N \leq \frac{3}{2}{p_1}^{\alpha_1}\sigma({p_1}^{\alpha_1})$
$\qquad B$ : $N$ is not an odd perfect number
$\qquad C$ : $\frac{\Omega(N) - 1}{2} < \alpha_1$
From $(1)$, we have
$$A\implies B$$
From $(2)$, we have
$$A\implies C$$
Now you are claiming in $(3)$ that
$$C\implies B$$
However, I don't see any proof for this.
So, I think that you have not proven what you claimed in $(4)$. 
A: (This is only a partial answer.  This approach considers the largest prime factor of $N$ (instead of the smallest) when computing a bound for $\Omega(N)$.)
If $$N = \prod_{i=1}^{\omega(N)}{{p_i}^{\alpha_i}}$$
${\it is}$ an odd perfect number (where the $p_i$'s are primes ordered in increasing magnitude and the $\alpha_i$'s are all positive), then  
$$N \geq \frac{3}{2}{p_i}^{\alpha_i}\sigma({p_i}^{\alpha_i}).$$
(See Theorem 4.2.5, page 112 in this M.Sc. thesis.)
(Let $r=\omega(N)$.)  In particular, suppose $i = r$.  (That is, let $p_r$ be the largest prime factor of $N$.)  Then we have
$${{p_r}^{\alpha_r}}\prod_{i=1}^{r-1}{{p_i}^{\alpha_i}} = N = \prod_{i=1}^{\omega(N)}{{p_i}^{\alpha_i}} \geq \frac{3}{2}{p_r}^{\alpha_r}\sigma({p_r}^{\alpha_r}) > \frac{3}{2}{{p_r}^{2\alpha_r}},$$
from which it follows that
$$\prod_{i=1}^{r-1}{{p_{r-1}}^{\alpha_i}} \geq \prod_{i=1}^{r-1}{{p_i}^{\alpha_i}} > \frac{3}{2}{{p_r}^{\alpha_r}}.$$
But we also have
$${p_r}^{\Omega(N) - {\alpha_r}} > {p_{r-1}}^{\Omega(N) - {\alpha_r}} = \prod_{i=1}^{r-1}{{p_{r-1}}^{\alpha_i}} \geq \prod_{i=1}^{r-1}{{p_i}^{\alpha_i}} > \frac{3}{2}{{p_r}^{\alpha_r}} > {p_r}^{\alpha_r},$$
from which we obtain
$$\Omega(N) > 2\alpha_r.$$
