Question Number 26007 by Tinkutara last updated on 17/Dec/17
Answered by ajfour last updated on 17/Dec/17
$${r}_{\mathrm{1}} \left(\mathrm{tan}\:\frac{{B}}{\mathrm{2}}+\mathrm{tan}\:\frac{{C}}{\mathrm{2}}\right)={a} \\ $$$${r}\left(\mathrm{cot}\:\frac{{B}}{\mathrm{2}}+\mathrm{cot}\:\frac{{C}}{\mathrm{2}}\right)={a} \\ $$$$\Rightarrow\:\frac{{r}_{\mathrm{1}} }{{r}}=\frac{\mathrm{cot}\:\frac{{B}}{\mathrm{2}}+\mathrm{cot}\:\frac{{C}}{\mathrm{2}}}{\mathrm{tan}\:\frac{{B}}{\mathrm{2}}+\mathrm{tan}\:\frac{{C}}{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{tan}\:\frac{{B}}{\mathrm{2}}\mathrm{tan}\:\frac{{C}}{\mathrm{2}}} \\ $$$${and}\:{we}\:{know}\:\Sigma\mathrm{tan}\:\frac{{B}}{\mathrm{2}}\mathrm{tan}\:\frac{{C}}{\mathrm{2}}=\mathrm{1} \\ $$$${let}\:{x}=\mathrm{tan}\:\frac{{A}}{\mathrm{2}}\:,\:{y}=\mathrm{tan}\:\frac{{B}}{\mathrm{2}},\:{z}=\mathrm{tan}\:\frac{{C}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\Sigma\left(\frac{{r}_{\mathrm{1}} }{{r}}\right)=\Sigma\frac{\mathrm{1}}{{xy}}\:\:{and}\:\:\Sigma{xy}=\mathrm{1} \\ $$$$\Rightarrow\:\:\:\Sigma\left(\frac{{r}_{\mathrm{1}} }{{r}}\right)\:\geqslant\:\mathrm{9}\:. \\ $$
Commented by Tinkutara last updated on 18/Dec/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir}! \\ $$