**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


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!



Proving a specific case of Robin's Inequality
By Theorem 1.2 in [this paper](https://jtnb.centre-mersenne.org/item/?id=JTNB_2007__19_2_357_0), Robin's Inequality is true for every odd integer $n>10$. If we knew what the OP wants to prove, then we would also know Robin's Inequality for every integer $n$ whose odd part exceeds $5040$. In particular, we would know Robin's Inequality for every [colossally abundant number](https://en.wikipedia.org/wiki/Colossally_abundant_number) exceeding $5040$, because each colossally abundant number divides the second next one (cf. Proposition 4 in [this paper](https://users.renyi.hu/~p_erdos/1975-37.pdf)). So, by Proposition 1 in Section 3 of [Robin's paper](http://zakuski.utsa.edu/~jagy/Robin_1984.pdf), we would know Robin's Inequality for every integer exceeding $5040$, which is equivalent to the Riemann Hypothesis.


In short, it is hopeless to prove what the OP wants to prove, because it implies the Riemann Hypothesis.


