Question Number 174837 by mnjuly1970 last updated on 12/Aug/22
$$ \\ $$$$\:\:\:\:\:{sin}\left({A}\right)+\:{sin}\left({B}\:\right)+\:{sin}\left({C}\right)\leqslant\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\succ\:\:\:{Solution}\:\prec\:\:\:\:\:\:\left(\:{A}+{B}\:+{C}\:=\pi\:\right) \\ $$$$\:\:\:\:\:\:{l}.{h}.{s}\:=\:\mathrm{2}{sin}\left(\frac{{A}+{B}}{\mathrm{2}}\:\right){cos}\left(\frac{{A}−{B}}{\mathrm{2}}\right)+\mathrm{2}{sin}\left(\frac{{C}}{\mathrm{2}}\right){cos}\left(\frac{{C}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}{cos}\:\left(\frac{{C}}{\mathrm{2}}\right)\left\{\mathrm{2}{cos}\frac{{A}}{\mathrm{2}}\:.{cos}\left(\frac{{B}}{\mathrm{2}}\:\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{4}{cos}\left(\frac{{A}}{\mathrm{2}}\right).{cos}\left(\frac{{B}}{\mathrm{2}}\right).{cos}\left(\frac{{C}}{\mathrm{2}}\right)\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:{l}.{h}.{s}\:\underset{{post}} {\overset{{previoue}} {\leqslant}}\:\mathrm{4}\:\left(\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}}\:\right)=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:{note}:\:\begin{cases}{\:{I}\::\:\:{cos}\left({a}\right)+{cos}\left({b}\right)=\mathrm{2}{cos}\left(\frac{{a}+{b}}{\mathrm{2}}\right){cos}\left(\frac{{a}−{b}}{\mathrm{2}}\right)}\\{\:{II}:\:{sin}\left({a}\right)+{sin}\left({b}\right)=\mathrm{2}{sin}\left(\frac{{a}+{b}}{\mathrm{2}}\right){cos}\left(\frac{{a}−{b}}{\mathrm{2}}\:\right)}\\{\:\:\:\begin{cases}{\:{cos}\left(\frac{\pi}{\mathrm{2}}\:−\alpha\right)={sin}\left(\alpha\right)}\\{\:\:{sin}\left(\frac{\pi}{\mathrm{2}}\:−\alpha\right)=\:{cos}\left(\alpha\right)}\end{cases}}\end{cases} \\ $$