First of all, I imagine it will not be correct, just because of its simplicity, but I would also want to know why, as I can't find any mistake on it.
The "proof" would be based on convining two main theorems/formulae. The first one, would be this one , due to Nicolas, where it is stated that RH would hold iff:
$$\frac{N_k}{\phi{(N_k)}} > e^{\gamma} \ln{(\ln{(N_k)})}$$
holded for every $k$ being $N_k$ the primorial of order $k$ and $\phi{(N_k)}$ its Euler's Totient function.
Then, my main aim here will be to prove that formula for every $k$. To do that, I will use this other theorem:
$$ \prod_{p \le n} \frac{p}{p-1} > e^{\gamma}(\ln{n})(1-\frac{1}{2(\ln{n})^{2}})$$
Taken from "Approximate Formulas for Some Functions of Prime Numbers" (link), Theorem 8 (3.28).
As, in this case, $\frac{N_k}{\phi{(N_k)}} = \prod_{p \le p_k} \frac{p}{p-1}$, we can try to see if this holds:
$$e^{\gamma}(\ln{p_k})(1-\frac{1}{2(\ln{p_k})^{2}})>e^{\gamma} \ln{\ln{N_k}}$$
Hence
$$\ln{p_k}-\frac{1}{2\ln{p_k}}>\ln{\ln{N_k}}$$
For it to be more clear, we can change $\ln{N_k}$ by $\theta{(k)}$ (Chebyshev's First Function) so that
$$\ln{p_k}-\frac{1}{2\ln{p_k}}>\ln{\theta{(k)}}$$
From there, we could easily get to
$$\frac{1}{2\ln{p_k}}<\ln{\frac{p_k}{\theta{(k)}}}$$
And, with the bounds of Theorems 3 (3.12) and 4 (3.15), we get
$$\frac{1}{2\ln{p_k}}<\ln{\frac{\ln{k}}{1+ \frac{1}{2\ln{k}}}}$$
What would be true for every big enought k, meaning that
$$\frac{N_k}{\phi{(N_k)}} > e^{\gamma} \ln{(\ln{(N_k)})}$$
holds, and, with so, RH.
Is this correct? Why would/would not it prove the RH?
Thank you!
Edit thanks to Jyrki Lahtonen
With $p_1, p_2, \ldots$ being a list of primes in increasing order we have $N_k = p_1 p_2 \cdots p_k$ for the primorial. Therefore $$\frac{N_k}{\phi(N_k)}=\prod_{p\le p_k}\frac{p}{p-1}.$$
Hence the lower bound is only $$ \frac{N_k}{\phi(N_k)} > e^\gamma \log p_k \left(1 - \frac1{2 \log^2 p_k}\right), $$ which does not work for the remaining argument.