**Edit:** It turns out that this is equivalent to the RH which gives the idea that this might a *a little* difficult to show. As such we could consider an even simpler case in which the number $n$ is squarefree (all values $k_j$ are equal to $1$. In previous papers it has been shown that squarefree numbers satisfy Robin's Inequality, but is this still the case for $2^kn$? If we make this loose condition we find our simpler form of 
$$
\dfrac{2^{k+1}-1}{2^{k}}\prod_{j=1}^m\dfrac{p_j+1}{p_j} < e^\gamma\log\log\left(2^k\prod_{j=1}^m p_j\right)
$$
with a minima of $f(k)$ at 
$$
k_{min} = -\dfrac{W_{-1}\left(-e^\gamma\prod_{j=1}^m \frac{1}{p_j+1}\right)+\log\left(\prod_{j=1}^m p_j\right)}{\log2}
$$
if we plug this in to our inequality we need to show that 
$$
e^\gamma\log\left(-W\left(-e^\gamma\prod_{j=1}^m\frac{1}{p_j+1}\right)\right) > \dfrac{e^{-W\left(-e^\gamma\prod_{j=1}^m\frac{1}{p_j+1}\right)-\log\left(\prod_{j=1}^m p_j\right)+1}-1}{e^{-W\left(-e^\gamma\prod_{j=1}^m\frac{1}{p_j+1}\right)-\log\left(\prod_{j=1}^m p_j\right)}}\prod_{j=1}^m\dfrac{p_j+1}{p_j}
$$


which I will admit is disgustingly messy, but looks (at least naively to me) potentially tractable since prime product and inverse prime product series are well studied.

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One reformulation of the Riemann Hypothesis is Robin's Inequality which states that for $n>5040$ the following holds iff the RH holds:
$$
\sigma(n) < e^\gamma n\log\log(n)
$$
where $\sigma$ is the sum of divisors function and $\gamma$ is the Euler Mascheroni Constant. Now for my specific case I want to show that given given some number $n=p_1^{k_1}p_2^{k_2}\ldots p_m^{m} > 5040$ where $p_j \neq 2$ is a prime number, if Robin's Inequality holds for $n$, then it must also hold for $2^k\cdot n$. Performing some algebra on the inequality we can see that this is the same as showing that if the following inequality holds
$$
\prod_{j=1}^m\dfrac{p_j^{k_j+1}-1}{p_j^{k_j}(p_j-1)} < e^\gamma\log\log\left(\prod_{j=1}^m p_j^{k_j}\right)
$$
then this inequality must hold as well
$$
\dfrac{2^{k+1}-1}{2^{k}}\prod_{j=1}^m\dfrac{p_j^{k_j+1}-1}{p_j^{k_j}(p_j-1)} < e^\gamma\log\log\left(2^k\prod_{j=1}^m p_j^{k_j}\right)
$$

I'll admit that I am not well versed in Analytic Number Theory, so this might be obvious and I have no idea, but so far I have only been able to show three fairly trivial things

1. According to numerical computations this seems to hold true. For large values of $n$ it appears that the R.H.S. is strictly larger that $2$ times the L.H.S. in the assumed inequality. 

2. Since the left side is bounded with respect to $k$ and the right side is not, there must exist some $N$ for which if $k\geq N$ then the inequality holds. Therefore there are only finite cases for which this inequality may not hold. 

3. Taking the difference of the left and right sides as this
$$
f(k) = e^\gamma\log\log\left(2^k\prod_{j=1}^m p_j^{k_j}\right) - \dfrac{2^{k+1}-1}{2^{k}}\prod_{j=1}^m\dfrac{p_j^{k_j+1}-1}{p_j^{k_j}(p_j-1)}
$$
has a derivative where $f'(0) < 0$ and $f'(N) > 0$, as such there exists a local minima of $f(k)$ which we can find to be the following value
$$
k_{min} = -\dfrac{W_{-1}\left(-e^\gamma\prod_{j=1}^m \frac{p_j-1}{p_j^{k_j+1}-1}\right)+\log\left(\prod_{j=1}^m p_j^{k_j}\right)}{\log2}
$$

where $W_{-1}$ is the second, more negative, solution of the Lambert W function when the argument is between $0$ and $-\frac{1}{e}$. The derivative also appears to be strictly positive past $k_{min}$ as 
$$
f'(k) = \dfrac{e^\gamma \log(2)}{k\log 2 + \log\left(\prod_{j=1}^m p_j^{k_j}\right)} - \dfrac{\log(2)}{2^k}\prod_{j=1}^m\dfrac{p_j^{k_j+1}-1}{p_j^{k_j}(p_j-1)}
$$
as this will behave like $\frac{1}{k} - \frac{1}{2^k}$ where the negative part decreases significantly faster than the positive part. 

If anyone could offer some potential insight that would be much appreciated!