On a criterion for almost perfect numbers using the abundancy index Let $\sigma(x)$ denote the sum of the divisors of $x$, and let $I(x) = \sigma(x)/x$ be the abundancy index of $x$.  A number $y$ is said to be almost perfect if $\sigma(y) = 2y - 1$.
In a preprint titled A Criterion For Almost Perfect Numbers Using The Abundancy Index, Dagal and Dris show that $N$ is almost perfect if and only if
$$\dfrac{2N}{N + 1} \leq I(N) < \dfrac{2N + 1}{N + 1}.$$
Here is my question:

Are these bounds for $I(N)$ (when $N$ is almost perfect) best-possible?

UPDATE (September 27 2016)

In an answer below the fold, I was able to obtain the improved upper bound
  $$I(N) < \dfrac{4N}{2N+1}.$$
  Can we likewise improve on the lower bound?

 A: Unfortunately the preprint linked in the OP appears to be quite poorly written, unless I've made a sign error somewhere which is always possible.  The fundamental premise of the paper is to give a condition on $I(n) = \sigma(n)/n$ that is necessary and sufficient for $n$ to be almost-perfect.
The necessary inequality will always be trivial, since every almost-perfect number satisfies $I(n) = 2 - \frac1n$, so one need only establish that this exact value lies between the two proposed bounds.  These are just rational linear functions of $n$ so the truth value of any comparison will be eventually constant: number theory cannot play any role here.
This leaves the sufficient condition as the only potential object of study.  But the strength of the sufficiency statement grows with the size of the interval.  To make the claim stronger we'd need to widen the bounds, not make them tighter.  This means that Lemma 2.1 in the paper (with an upper bound of $2$) is actually stronger than the main theorem, which the authors are fundamentally misguided in not recognizing.
I'm afraid this question is similarly misguided: there is nothing whatsoever to be gained by decreasing the upper bound: we already know the exact upper bound is $(2n-1)/n$, so it's trivial to make the upper bound arbitrarily close to this by taking averages or mediants.  Take the mediant of your bound $4n/(2n+1)$ and the true value $(2n-1)/n$: voilà, you get an even smaller upper bound of $(6n-1)/(3n+1)$.  This game can be played endlessly because we already know the exact value of the abundancy of almost-perfect numbers.
Even the game of increasing the upper bound is rather pointless.  If we make the plausible (yet hopeless to prove) assumption that there are infinitely many perfect numbers, then there is no hope of increasing the upper bound beyond $2$.  The upper bound becomes a necessarily-trivial matter of finding any rational function that lies between $2-\frac1n$ and $2$.  The given bound of $(2n+1)/(n+1)$ is just the mediant of the two fractions $(2n-1)/n$ and $2/1$, which obviously lies between them
with no need for calculation.
No doubt similar tricks can be played with the lower bound, but it's not immediately obvious to me that there should be infinitely many numbers satisfying $\sigma(n) = 2n-2$.
A: It turns out that we can improve on the upper bound
$$I(N) < \dfrac{2N + 1}{N + 1} = \dfrac{2N + D(N)}{N + D(N)},$$
where $2N-\sigma(N)=D(N)$ is the deficiency of $N$.
To this end, consider the number
$$\dfrac{2N}{N + \dfrac{D(N)}{2}}.$$
First, we show that a number is almost perfect if and only if
$$I(N) < \dfrac{2N}{N + \dfrac{D(N)}{2}}$$
where $D(N)$ is of course equal to $1$.  (We will later need an improved criterion for $D(N)>1$, so we will leave the notation as is.)

CLAIM
  If $D(N) \geq 1$, then
  $$I(N) < \dfrac{4N}{2N + D(N)}.$$

Suppose to the contrary that
$$I(N) \geq \dfrac{4N}{2N + D(N)}.$$
Since
$$I(N) = \dfrac{\sigma(N)}{N} = \dfrac{2N - D(N)}{N}$$
we have
$$\dfrac{2N - D(N)}{N} = I(N) \geq \dfrac{4N}{2N + D(N)}$$ 
$$\iff 4N^2 - (D(N))^2 = \left(2N - D(N)\right)\cdot\left(2N + D(N)\right) \geq 4N^2$$
(where we have used $2N - D(N) = \sigma(N) > 0$)
$$\iff -(D(N))^2 \geq 0$$
This contradicts $D(N) \geq 1$.  Hence, our CLAIM is proved.
Next, we prove that
$$\frac{4N}{2N + D(N)} < \frac{2N + D(N)}{N + D(N)}.$$
Assume to the contrary that
$$\frac{2N + D(N)}{N + D(N)} \leq \frac{4N}{2N + D(N)}.$$
Since $D(N) \geq 1$, we obtain
$$\frac{2N + D(N)}{N + D(N)} \leq \frac{4N}{2N + D(N)} \iff \left(2N + D(N)\right)^2 \leq {4N}\cdot\left(N + D(N)\right)$$
$$\iff 4N^2 + {4N}\cdot{D(N)} + (D(N))^2 \leq 4N^2 + {4N}\cdot{D(N)} \iff (D(N))^2 \leq 0.$$
This last inequality contradicts $D(N) \geq 1$.  It follows that
$$\frac{4N}{2N + D(N)} < \frac{2N + D(N)}{N + D(N)}.$$
CONCLUSION

$4N/(2N + D(N))=4N/(2N + 1)$ is an improved upper bound for $I(N)$ when $D(N)=1$ (i.e., when $N$ is almost perfect).

In general, we have the following THEOREM.
THEOREM

If $2N-\sigma(N)=D(N) \geq 1$, then
  $$\frac{2N}{N+D(N)} \leq I(N)<\frac{4N}{2N+D(N)}.$$
  Equality holds when $D(N)=N=1$.

