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!
 A: Let $a=x-y$, $b=y-z$, $c=z-x$, then $a+b+c=0$ and there is no loss in generality.
This gives, using trigonometric identities
$$
F = \left|\sin{a}+\sin{b}+\sin{c}\right| \\
= \left|\sin{(x-y)}+\sin{(y-z)}+\sin{(z-x)}\right|\\
= 4  \left|\sin(x/2 - y/2) \sin(x/2 - z/2) \sin(y/2 - z/2)\right|
$$
Now the last line shows that, whatever the choice $(x,y,z)$ is which maximizes this, we can add a constant to all three values in this choice $(x,y,z)$ without changing anything. 
So let's put $x=0$, $y = m+d$, $z = m-d$ [m for mean, d for difference] which gives, using again trigonometric identities,
$$
F 
= 4  \left|\sin((m+d)/2) \sin((m-d)/2) \sin(d)\right| \\
= 2 \left|\sin(d) \right| \left| \cos(d) - \cos(m)\right| 
$$
Now for $\cos(d) > 0$, $m$ can be adjusted to $-\pi$ (for $\cos(d) < 0$, adjust $m=0$ which gives the same result) such that F is maximized to be (writing for both cases of $\cos(d) >< 0$)
$$
F \le
  2 \left|\sin(d) \right| (\left| \cos(d) \right| +1)
$$
Put $\left| \cos(d) \right|= q$ then 
$$
F \le
 2 \sqrt{1-q^2} (q +1)
$$
which has a maximum at $F \le \frac{3\sqrt{3}}{2}$, as required. The maximum occurs at  $q^* = |\cos(d^*)| = 0.5 $. For $\cos(d^*) = 0.5$, this is  $d^* = 2 \pi n + \pi/3$, and for $\cos(d^*) = -0.5$, this is  $d^* = 2 \pi n + 2 \pi/3$, both giving the same maximum value for $F$.
A: By the triangle inequality, C-S and AM-GM we obtain:
$$|\sin{a}+\sin{b}+\sin{c}|=|\sin{a}+\sin{b}-\sin{a}\cos{b}-\cos{a}\sin{b}|=$$
$$=|(1-\cos{b})\sin{a}-\sin{b}\cos{a}+\sin{b}|\leq|(1-\cos{b})\sin{a}-\sin{b}\cos{a}|+|\sin{b}|\leq$$
$$\leq\sqrt{((1-\cos{b})^2+(-\sin b)^2)(\sin^2a+\cos^2a)}+|\sin{b}|=$$
$$=\sqrt{2-2\cos{b}}+|\sin{b}|=2|\sin\frac{b}{2}|+|\sin{b}|=2|\sin\frac{b}{2}|\left(1+|\cos\frac{b}{2}|\right)=$$
$$=2\sqrt{\left(1-|\cos\frac{b}{2}|\right)\left(1+|\cos\frac{b}{2}|\right)^3}=\frac{2}{\sqrt3}\sqrt{\left(3-3|\cos\frac{b}{2}|\right)\left(1+|\cos\frac{b}{2}|\right)^3}\leq$$
$$\leq\frac{2}{\sqrt3}\sqrt{\left(\frac{3-3|\cos\frac{b}{2}|+3\left(1+|\cos\frac{b}{2}|\right)}{4}\right)^4}=\frac{3\sqrt3}{2}.$$
