**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.

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

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.

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.

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!