Question Number 122636 by mnjuly1970 last updated on 18/Nov/20
$$\:\:\:\:\:\:\:\:\:\:…{nice}\:\:{calculus}… \\ $$$$\:\:\:{In}\:\:\mathrm{A}\overset{\Delta} {\mathrm{B}C}\:\:{prove}\:::\: \\ $$$$\:\:\:\ast:\:\:{sin}\left(\frac{\mathrm{A}}{\mathrm{2}}\right){sin}\left(\frac{\mathrm{B}}{\mathrm{2}}\right){sin}\left(\frac{\mathrm{C}}{\mathrm{2}}\right)\leqslant\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$……………………. \\ $$$$\:\:\:\:\ast\ast::\:\:\:{max}\left({cos}\left(\frac{\mathrm{A}}{\mathrm{2}}\right){cos}\left(\frac{\mathrm{B}}{\mathrm{2}}\right){cos}\left(\frac{\mathrm{C}}{\mathrm{2}}\right)\right)=? \\ $$$$\:\:\:\:\: \\ $$$$\: \\ $$$$\:\:\:\:\:\:\: \\ $$
Answered by TANMAY PANACEA last updated on 18/Nov/20
$$\:{three}\:{points}\:{A},{B}\:{and}\:{C}\:{lie}\:{on}\:{curve}\:{y}={sin}\frac{{x}}{\mathrm{2}} \\ $$$$\left({A},{sin}\frac{{A}}{\mathrm{2}}\right)\:\:\left({B},{sin}\frac{{B}}{\mathrm{2}}\right){and}\:\left({C},{sin}\frac{{C}}{\mathrm{2}}\right) \\ $$$${centroid}\:{P}\:\left(\frac{{A}+{B}+{C}}{\mathrm{3}},\frac{{sin}\frac{{A}}{\mathrm{2}}+{sin}\frac{{B}}{\mathrm{2}}+{sin}\frac{{C}}{\mathrm{2}}}{\mathrm{3}}\right) \\ $$$${ordinate}\:{of}\:{centroid}\:{P}=\left(\frac{{sin}\frac{{A}}{\mathrm{2}}+{sin}\frac{{B}}{\mathrm{2}}+{sin}\frac{{C}}{\mathrm{2}}}{\mathrm{3}}\right) \\ $$$${pointQ}=\left\{\:\left(\frac{{A}+{B}+{C}}{\mathrm{3}}\right){sin}\left(\frac{\frac{{A}+{B}+{C}}{\mathrm{3}}}{\mathrm{2}}\right)\right\}{lies} \\ $$$${on}\:{y}={sin}\left(\frac{{x}}{\mathrm{2}}\right) \\ $$$${sin}\left(\frac{{A}+{B}+{C}}{\mathrm{6}}\right)={ordinate}\:{of}\:{pointQ} \\ $$$${sin}\left(\frac{{A}+{B}+{C}}{\mathrm{6}}\right)>\frac{{sin}\frac{{A}}{\mathrm{2}}+{sin}\frac{{B}}{\mathrm{2}}+{sin}\frac{{C}}{\mathrm{2}}}{\mathrm{3}} \\ $$$${using}\:{AM}>{GM} \\ $$$${sin}\left(\frac{\mathrm{180}^{{o}} }{\mathrm{6}}\right)>\frac{{sin}\frac{{A}}{\mathrm{2}}+{sin}\frac{{B}}{\mathrm{2}}+{sin}\frac{{C}}{\mathrm{2}}}{\mathrm{3}}>\left({sin}\frac{{A}}{\mathrm{2}}{sin}\frac{{B}}{\mathrm{2}}{sin}\frac{{C}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} >\left({sin}\frac{{A}}{\mathrm{2}}{sin}\frac{{B}}{\mathrm{2}}{sin}\frac{{C}}{\mathrm{2}}\right) \\ $$$${proved} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by TANMAY PANACEA last updated on 18/Nov/20
$${same}\:{method}\:{for}\:{cos}\theta\:{curve} \\ $$$${final}\:{step} \\ $$$${cos}\left(\frac{\mathrm{180}^{{o}} }{\mathrm{6}}\right)>\left({cos}\frac{{A}}{\mathrm{2}}{cos}\frac{{B}}{\mathrm{2}}{cos}\frac{{C}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{3}} >\left({cos}\frac{{A}}{\mathrm{2}}{cos}\frac{{B}}{\mathrm{2}}{cos}\frac{{C}}{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}>\left({cos}\frac{{A}}{\mathrm{2}}{cos}\frac{{B}}{\mathrm{2}}{cos}\frac{{C}}{\mathrm{2}}\right) \\ $$
Commented by mnjuly1970 last updated on 18/Nov/20
$${thank}\:{you}\:{sir}\:{tanmay}… \\ $$$$\:\:{excellent}… \\ $$
Commented by TANMAY PANACEA last updated on 18/Nov/20
$${most}\:{welcome} \\ $$