On an oscillation Theorem involving the Chebyshev function and the zeros of the Riemann zeta function Define $\theta(x)=\sum_{p\leq x} \log p $, where $p>1$ denotes a prime.
Nicolas proved that if the Riemann zeta function $\zeta(s)$ vanishes for some $s$ with $\Re(s)\leq 1/2 + b$, where $b\in(0, 1/2)$, then 
$$\log\Big(e^{\gamma}(\log \theta(x))\prod_{p\leq x} (1-p^{-1})\Big)=\Omega_{\pm }(x^{-b}).$$
Is the reverse necessarily true ? It appears so to me, since
$$\theta(x)=x +O(\sqrt{x})-\sum_{\rho} \frac{x^{\rho}}{\rho},$$
where the sum is over the complex zeros of $\zeta$ and $\prod_{p\leq x} (1-p^{-1})\sim e^{-\gamma}(\log x)^{-1} ?$
 A: The result Nicolas proved (Theorem 3, Jean-Louis Nicolas. Petites valeurs de la fonction d'Euler, J. Number Theory, 17, 1983, 375--388, paper linked HERE) is actually not quite what is claimed in this post, but:
$\log\Big(e^{\gamma}(\log \theta(x))\prod_{p\leq x} (1-p^{-1})\Big)=\Omega_{\pm }(x^{b-\frac{1}{2}-\epsilon})$, for all $\epsilon > 0$, where $0< b < \frac{1}{2}$ is such that RZ has a zero with real part $\frac{1}{2} + b$, so it is in the classical spirit as the worse RH fails (the higher the supremum of the real part of the critical zeros), the stronger the claimed $\Omega$. 
The estimate claimed in the post above is wrong (obviously because it is stronger for a weaker claim, not to speak that since RZ always vanishes on the critical line the claim of the post would imply the estimate to be true for $b$ going to zero and that contradicts the later claim when RZ is true). Using the zeros symmetry around the critical line, one can rephrase the $\Omega$ result to:
$\log\Big(e^{\gamma}(\log \theta(x))\prod_{p\leq x} (1-p^{-1})\Big)=\Omega_{\pm }(x^{-\beta-\epsilon})$, for all $\epsilon > 0$, where $0< \beta < \frac{1}{2}$ is such that RZ has a zero with real part $\beta$
He also proved that if RH is true $\log\Big(e^{\gamma}(\log \theta(x))\prod_{p\leq x} (1-p^{-1})\Big)x^{\frac{1}{2}}\log x$ is always negative for $x>2$, has inferior limit $\log{4\pi}-4-\gamma$ and superior limit less or equal to $\gamma - \log{4\pi}$ which are both close to $-2$, so obviously any $\Omega$ result as above with a power $x^{-\beta}, 0 < \beta < \frac{1}{2}$ dispproves RH, with the lower the $\beta$ the worse RH fails, while any $O$ result gives bounds on the supremum of the real parts of critical zeros, with the higher the $\beta$ the better the bounds
A: Actually, in the paper cited in answer one, the following is proved:
$\log (e^\gamma\log(\theta(x))\prod_{p\le x}(1-1/p))<0$ for $x\ge 3$ if and only if the Riemann Hypothesis holds ... I don't know how to link to other posts, but essentially this same question is asked again (this one came first).
