For the "greater-than" problem in Yao's 2-party communication complexity model, Alice receives $X$ and Bob receives $Y$, and they need to decide whether $X>Y$.
I recently listened to an (online) seminar talk where the speaker mentioned that it is known that any randomized $k$-round protocol that computes "greater-than" requires $\Omega\left(\frac{(\log n)^{1/k}}{k^2}\right)$ bits.
Unfortunately, I didn't get to ask the speaker about where this result appeared and I didn't have any luck searching for it online. Is there a reference where I could find the proof for this round-communication trade-off?