<s>[INST] Prove a trigonometric inequality Let $a,b,c$ reals with sum 0. Prove that
$\left|\sin{a}+\sin{b}+\sin{c}\right| \leq \frac{3\sqrt{3}}{2}.$
My idea was to set $z_1=\cos{a}+i\sin{a}$ and also $z_2,z_3$. Then I tried replacing the sines in the inequality with these complex numbers. However, I didn’t get to anything interesting that could help me prove this.
Thanks in advance!
 [/INST] By the triangle inequality, C-S and AM-GM we obtain:
$$|\sin{a}+\sin{b}+\sin{c}|=|\sin{a}+\sin{b}-\sin{a}\cos{b}-\cos{a}\sin{b}|=$$
$$=|(1-\cos{b})\sin{a}-\sin{b}\cos{a}+\sin{b}|\leq|(1-\cos{b})\sin{a}-\sin{b}\cos{a}|+|\sin{b}|\leq$$
$$\leq\sqrt{((1-\cos{b})^2+(-\sin b)^2)(\sin^2a+\cos^2a)}+|\sin{b}|=$$
$$=\sqrt{2-2\cos{b}}+|\sin{b}|=2|\sin\frac{b}{2}|+|\sin{b}|=2|\sin\frac{b}{2}|\left(1+|\cos\frac{b}{2}|\right)=$$
$$=2\sqrt{\left(1-|\cos\frac{b}{2}|\right)\left(1+|\cos\frac{b}{2}|\right)^3}=\frac{2}{\sqrt3}\sqrt{\left(3-3|\cos\frac{b}{2}|\right)\left(1+|\cos\frac{b}{2}|\right)^3}\leq$$
$$\leq\frac{2}{\sqrt3}\sqrt{\left(\frac{3-3|\cos\frac{b}{2}|+3\left(1+|\cos\frac{b}{2}|\right)}{4}\right)^4}=\frac{3\sqrt3}{2}.$$
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