Note that if $1/2<\sigma<1, t \ne 0, \rho=\sigma+it$ we have $|\rho|^2=\sigma^2+t^2>(1-\sigma)^2+t^2=|1-\rho|^2$ so $\frac{2\sigma -1}{|\rho|^2} < \frac{2\sigma -1}{|1-\rho|^2}$ (since $2\sigma-1>0$) and by rearranging a little we get:
$\frac{2\sigma}{|\rho|^2} +\frac{2(1-\sigma)}{|1-\rho|^2}< \frac{1}{|\rho|^2}+\frac{1}{|1-\rho|^2}$
But now the expression above is symmetric in $\sigma, 1-\sigma$ so by interchanging them the inequality holds for $\sigma < 1/2$ too
Assume $\sum_\rho\frac{1}{\rho(1-\rho)}= \sum_{\rho}\frac{1}{|\rho|^2}$ and note that both sides are absolutely convergent hence we can regroup at will.
In particular, we rearrange the terms so we group $\rho, 1-\rho, \bar \rho, 1-\bar \rho$ together. If $\Re \rho =1/2$ this grouping has only two terms and appears twice (once for $\rho$ and once for $1-\rho$) while the sum is $\frac{1}{|\rho|^2}=\frac{1}{|1-\rho|^2}$ so the corresponding terms on RHS are equal.
If RH is false there is $\rho=\sigma+it, \sigma \ne 1/2, t > 0$ and the grouping above has $4$ terms each appearing twice (this time for $\rho, 1-\bar \rho$) with sum $\frac{1}{\rho(1-\rho)}+\frac{1}{\bar \rho(1-\bar \rho)}= \frac{2\sigma}{|\rho|^2} +\frac{2(1-\sigma)}{|1-\rho|^2}$ However in this case the corresponding terms on RHS are $\frac{1}{|\rho|^2}+\frac{1}{|1-\rho|^2}$ and by the result in the beginning we get a strict inequality.
Since by the grouping above we showed that LHS is a sum of positive terms, some equal with the corresponding ones on RHs and some strictly less we get the contradiction $\sum_\rho\frac{1}{\rho(1-\rho)} < \sum_{\rho}\frac{1}{|\rho|^2}$ so our supposition that RH is false cannot hold. Done!
Best Answer
This is a consequence of partial summation. Note that $$\sum_{\rho} |\rho|^{-A} = \lim_{T \to \infty} \sum_{0 < Im(\rho) < T} |\rho|^{-A} = \lim_{T \to \infty} \int_0^T t^{-A} d N(t).$$ Integration by parts shows that this equals $$\lim_{T \to \infty} \frac{N(T)}{T^A} + A\int_0^T \frac{N(t)}{t^{A+1}} dt.$$ Since $N(T) \ll T \log T$, if $A > 1$ we have $$\frac{N(T)}{T^A} = O\left(\frac{\log T}{T^{A-1}}\right) \to 0$$ as $T \to \infty$ and $$\frac{N(t)}{t^{A+1}} = O\left(A \frac{\log t}{t^A}\right)$$ is integrable as $t \to \infty$. Hence $$\sum_{\rho} |\rho|^{-A} = A\int_0^{\infty} \frac{N(t)}{t^{A+1}} dt$$ is finite.
Note that $\sum_{\rho} |\rho|^{-1} = \infty$ does not follow from $N(T) \ll T \log T$ (Pretend for a moment that Riemann zeta function has finitely many zeros, then $N(T) \ll T \log T$ is of course true but $\sum_{\rho} |\rho|^{-1}$ converges). You would need $N(T) \asymp T \log T$ for that claim.