Finding minimum of $\max(\arg(a),\arg(b),\arg(c))$ in complex number 
If $a,b,c$ be $3$ complex number (not all real ) such that $|a|=|b|=|c|=1$ and $2(a+b+c)-3abc$ is real , Then minimum of $\max(\arg(a),\arg(b),\arg(c))$ Given argument of $a,b,c$ all are positive.

Try: Assuming $z =2(a+b+c)-3abc$. Then $z=\bar{z}$
So $$2(a+b+c)-3abc=2(\overline{a}+\overline{b}+\overline{c})-3\overline{abc}$$
Could some help me to solve it , thanks

We claim that the answer is $\dfrac{\pi}{6}$ and proceed to argue by contradiction.
Let $a = e^{ix}, b = e^{iy}, c = e^{iz}$ and assume to the contrary that there exists  $x,y,z \in [0, \dfrac{\pi}{6})$, such that  $$2(e^{ix}+e^{iy}+e^{iz})-3e^{i(x+y+z)} = 2(e^{-ix}+e^{-iy}+e^{-iz})-3e^{-i(x+y+z)}$$ or $$2(\sin x+\sin y +\sin z) = 3\sin(x+y+z)\,\ \dagger. $$ 
However, we see that $t = x+y+z<\dfrac{\pi}{2}$ and on $(0, \dfrac{\pi}{2})$, we will prove the following inequality: 
$$\sin t>2\sin\dfrac{t}{3}\,\,\ddagger.$$
Consider $f(x) = \sin 3x  - 2\sin x = \sin x - 4\sin^3x = \sin x(1-2\sin x)(1+2\sin x)>0$ when $0<x<\pi/6$, so $\ddagger$ is proven.
Finally, since $\sin t$ is concave on our interval, we apply the Jensen's inequality and combine with $\ddagger $ to obtain:
$$3\sin(x+y+z)> 6\sin\left(\dfrac{x+y+z}{3}\right)\geq 2(\sin x+\sin y+\sin z) = 3\sin(x+y+z),$$
a contradiction.
Therefore, $\max\{x,y,z\}\geq\dfrac{\pi}{6}$ and the minimum of that max is actually $\pi/6$, when $x,y,z = \dfrac{\pi}{6}.$
