I was working on a problem recently, and it happened that it could be solved if $3R\le 2h_{\max}$ was true for all acute angled triangles. So I used GeoGebra to check it, and found that for some angles greater than $80^{\circ}$, it did not hold true. So although it did not provide a solution to my problem, it inspired this question:
In a triangle $ABC$, with $\angle A \ge \angle B \ge \angle C$, what may be the maximum value of $\angle A$ such that $3R\le 2h_{\max}$ holds true, where $R$ is the circumradius, and $h_{\max}$ is the largest altitude.
How can we solve this?
Given $\angle A\ge \angle B\ge\angle C$, $h_c$ is the largest attitude. So we want $$3R\le 2h_c=4R\sin A\sin B,$$ or $$\frac34\le \sin A\sin B.\tag{1}$$ If $A\sim B$, then they are both larger than $\frac\pi3$, and (1) is true.
On the other hand, if we want (1) to hold for all $\triangle ABC$, then we see that $\sin B\ge \sin\frac{\pi-A}{2}$, thus we want $$\frac34\le \sin A\sin\frac{\pi-A}{2}.\tag{2}$$ Let $t=\sin\frac{A}{2}$, then $\frac12\le t\le 1$ and (2) becomes $$\frac34\le 2t(1-t^2)$$ or $$8t^3-8t+3\le 0.\tag{3}$$ The RHS of (3) has three zeros: $\frac12$, $\frac{-1\pm\sqrt{13}}{4}$, which implies that $$t\le \frac{\sqrt{13}-1}{4}.$$ That means $$\angle A\le 2\arcsin\left(\frac{\sqrt{13}-1}{4}\right) (\sim 81.3^\circ).$$