Question Number 222638 by Mingma last updated on 03/Jul/25
Answered by gabthemathguy25 last updated on 03/Jul/25
$$\mathrm{cot}\:{A}\:+\:\mathrm{cot}\:{B}\:+\:\mathrm{cot}\:{C}\:=\:\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{4}{S}} \\ $$$${where}\:{a},{b},{c}\:=\:{side}\:{lengths},\:{S}={area}\:{of}\:{triangle} \\ $$$${S}={r}\centerdot{s}\:\Rightarrow\:\frac{\mathrm{1}}{{S}}=\frac{\mathrm{1}}{{rs}} \\ $$$${so}\:\:\:\:\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{4}{rs}} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{9}{R}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{4}{rR} \\ $$$${therefore}\:\frac{\mathrm{9}{R}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{4}{rR}}{\mathrm{4}{rs}} \\ $$$${s}\geq\mathrm{3}\sqrt{\mathrm{3}},\frac{\mathrm{1}}{{s}}\leq\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{3}}{r}} \\ $$$${so}\:\mathrm{cot}\:\:{A}\:+\:\mathrm{cot}\:{B}\:+\:\mathrm{cot}\:{C}\:\leq\:\frac{\mathrm{9}{R}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{4}{rR}}{\mathrm{4}{r}\centerdot\mathrm{3}\sqrt{\mathrm{3}}{r}}=\frac{\mathrm{9}{R}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{4}{rR}}{\mathrm{12}\sqrt{\mathrm{3}}{r}^{\mathrm{2}} } \\ $$$$\mathrm{cot}\:{A}\:+\:\mathrm{cot}\:{B}\:+\:\mathrm{cot}\:{C}\:<\frac{\mathrm{9}{R}^{\mathrm{2}} +{smaller}\:{terms}}{\mathrm{12}\sqrt{\mathrm{3}}{r}^{\mathrm{2}} }<\frac{\mathrm{9}{R}^{\mathrm{2}} }{\mathrm{12}\sqrt{\mathrm{3}}{r}^{\mathrm{2}} }=\frac{\mathrm{3}{R}^{\mathrm{2}} }{\mathrm{4}\sqrt{\mathrm{3}}{r}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{3}\sqrt{\mathrm{3}}{R}^{\mathrm{2}} }{\mathrm{4}\centerdot\mathrm{3}{r}^{\mathrm{2}} }=\frac{\sqrt{\mathrm{3}}{R}^{\mathrm{2}} }{\mathrm{4}{r}^{\mathrm{2}} }=\sqrt{\mathrm{3}}\left(\frac{{R}}{\mathrm{2}{r}}\right)^{\mathrm{2}} \\ $$