Courtney Griffin

2022-01-25

In any triangle is $\mathrm{sin}A+\mathrm{sin}B+\mathrm{sin}C=\frac{3\sqrt{3}}{2}$ always Well, I came with an interesting proof. But I just want to verify it Applied at function $y=\mathrm{sin}x$ We have, $\mathrm{sin}\left(\frac{A+B+C}{2}\right)\ge \frac{\mathrm{sin}A+\mathrm{sin}B+\mathrm{sin}C}{3}$ From here we will get $\mathrm{sin}A+\mathrm{sin}B+\mathrm{sin}C\le \frac{3\sqrt{3}}{2}$ Also by A.M. $\ge$ G.M. in an acute angled triangle $\frac{\mathrm{sin}A+\mathrm{sin}B+\mathrm{sin}C}{3}\ge \sqrt[3]{\mathrm{sin}A\mathrm{sin}B\mathrm{sin}C}$ $⇒\mathrm{sin}A+\mathrm{sin}B+\mathrm{sin}C\ge 3\left(\sqrt[3]{\mathrm{sin}A\mathrm{sin}B\mathrm{sin}C}\right)$ $⇒\mathrm{sin}A+\mathrm{sin}B+\mathrm{sin}C\ge 3\left(\frac{\sqrt{3}}{2}\right)=\frac{3\sqrt{3}}{2}>2$ and from this I get $\mathrm{sin}A+\mathrm{sin}B+\mathrm{sin}C\ge \frac{3\sqrt{3}}{2}$

Lily Thornton

Expert