Question Number 175956 by mnjuly1970 last updated on 10/Sep/22
Commented by mnjuly1970 last updated on 10/Sep/22
$${prove}\:\upuparrows\upuparrows \\ $$
Answered by behi834171 last updated on 10/Sep/22
$${sin}\frac{{A}}{\mathrm{2}}.{sin}\frac{{B}}{\mathrm{2}}.{sin}\frac{{C}}{\mathrm{2}}\leqslant{sin}^{\mathrm{3}} \frac{\frac{\pi}{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\Rightarrow\Sigma\frac{\mathrm{1}}{{sin}^{\mathrm{2}} \frac{{A}}{\mathrm{2}}}\geqslant\mathrm{3}×\sqrt[{\mathrm{3}}]{\Pi\frac{\mathrm{1}}{{sin}^{\mathrm{2}} \frac{{A}}{\mathrm{2}}}}\geqslant \\ $$$$\:\:\:\geqslant\mathrm{3}×\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{64}}}}=\mathrm{3}×\mathrm{4}=\mathrm{12}\:\:\:.\blacksquare \\ $$
Commented by mnjuly1970 last updated on 10/Sep/22
$${mamnoon}\:{ostad} \\ $$$${faghat}\::\:{sin}\left(\frac{{A}}{\mathrm{2}}\right){sin}\left(\frac{{B}}{\mathrm{2}}\right){sin}\left(\frac{{C}}{\mathrm{2}}\right)\leqslant\frac{\mathrm{1}}{\mathrm{8}} \\ $$
Commented by mnjuly1970 last updated on 10/Sep/22
$${zendeh}\:{bashid} \\ $$
Commented by behi834171 last updated on 10/Sep/22
$$\mathrm{3}{pas}\:{ostad}.{eslah}\:{shod}. \\ $$