Let $T>10$. Give a tighter lower bound of $T$ if $\log T / T \leq \epsilon^{2}$.
I know that $T$ should be in the order of $\widetilde{O}(\epsilon^{-2})$. How can we prove this?
Suppose $T = A \epsilon^{-2+\delta}$.
We then have $\frac{\log T \epsilon^{-2}}{T} = \frac{(-2+\delta) \log \epsilon + log A}{A\epsilon^{\delta}} \rightarrow \infty$ if $\delta>0$.
So $\delta$ must be zero. But this does not give a lower bound of $T.$
As @RiverLi commented $$\frac{\log(T)} T=\epsilon^2 \quad \implies \quad T=-\frac{W_{-1}\left(-\epsilon ^2\right)}{\epsilon ^2}$$ where $W$ is the second branch of Lambert function.
In year $2016$, Chatzigeorgiou proved that, if $u>0$,
$$-1 - \sqrt{2u} - u < W_{-1}\left(-e^{-u-1}\right) < -1 - \sqrt{2u} - \frac{2}{3}u$$
Let $u=-(2 \log (\epsilon )+1)$ to obtain two simple bounds for $T$