Big-O of Upperbound on the Regret of Exp3

Question

I'm having difficulty understanding how to compute Big-O for the upper bound on the regret in Exp3 algorithm. I think the actual algorithm isn't quite important for my question but since I couldn't find the relevant material to work out so let me shorten the detailed notations.

So, the following is the famous bound on the regret of Exp3.

Theorem(Expected regret of Exp3)

The expected regret of Exp3 with learning rate $\eta > 0$ is; $$ \mathcal{R}_T \leq \frac{\ln N}{\eta} + T \eta $$

And most of the materials(mostly lecture notes: here or here) that I went through are concluding the derivation of the above theorem by saying that if you set $\eta$ to be as follows, you get the following Big-O version of the upperbound. But how do we compute it?

Collorary

The expected regret of Exp3 with learning rate $\eta = \sqrt{\frac{ \ln N}{T}}$ is; $$ \mathcal{R}_T \leq O(\sqrt{T \ln N}) $$

Answer 1

The upper bound has the form $\frac{A}x+Bx$. This is in both terms and also in the sum a convex function over the positive half-axis, raising to infinity for $x\to 0$ and $x\to\infty$. So there is a unique minimum.

This minimum can be determined via calculus or via mean inequalities, $$ \frac{\frac{A}x+Bx}2\ge\sqrt{AB}, $$ with equality for $\frac{A}x=Bx\implies x=\sqrt{\frac{A}{B}}$.