An inequality involving complex numbers I would like to prove the following inequality:
$$\large{
\left| {1 + re^{ - \frac{2}{3}\theta i} } \right|^2 \left| {1 + re^{\frac{2}{3}\left( {\pi  - \theta } \right)i} } \right|^2  \ge \cos ^2 \theta ,
}$$
for any $r>0$ and $-\frac{\pi}{2}<\theta<\frac{3\pi}{2}$. I expanded the left-hand side as a polynomial in $r$ but it seems hopeless. Any idea? Thank you very much in advance!
 A: Unfortunately I haven't found a shorter proof so far.

We have
\begin{align}
f(r,\theta)&=\left|1+re^{-2i\theta/3}\right|^2\left|1+re^{2i(\pi-\theta)/3}\right|^2=\\&=\left(1+r^2+2r\cos\frac{2\theta}{3}\right)\left(1+r^2+2r\cos\frac{2(\pi-\theta)}{3}\right)=\\
&=\frac{3}{4}\left(1-r^2\right)^2+4r^2\left(\frac{1+r^2}{4r}+\cos\frac{\pi-2\theta}{3}\right)^2.
\end{align}
On the other hand,
$$\cos^2\theta=\frac{1+\cos2\theta}{2}=\frac{1-\cos\left(3\times\frac{\pi-2\theta}{3}\right)}{2}=\frac12\left(1-4\cos^3\frac{\pi-2\theta}{3}+3\cos\frac{\pi-2\theta}{3}\right).$$

Therefore, if we denote $q=\cos\frac{\pi-2\theta}{3}$, all we need to show is that for $r>0$ and $q\in(-\frac12,1]$ one has
$$\frac{3}{4}\left(1-r^2\right)^2+4r^2\left(\frac{1+r^2}{4r}+q\right)^2
\geq \frac12\left(1-4q^3+3q\right)\tag{1}$$
or, equivalently,
$$g(q,r)=4q^3+8r^2q^2+(4r^3+4r-3)q+(2r^4-2r^2+1)\geq0.\tag{2}$$

Now since it can be easily shown that
\begin{align}
&g(-1/2,r)=2(r-1)^2(r^2+r+1)\geq0,\\
&g(1,r)=2(r^2+r+1)^2>0,
\end{align}
the only possibility for (2) to be violated is to have $\displaystyle\frac{\partial g}{\partial q}=0$ at some point $q_{min}\in(-\frac12,1)$. Differentiating $g$ and solving the resulting quadratic equation for $q$, one can show that such point (+ giving a minimum) can exist only if $r\in(0,\frac34]\cup[(\frac34)^{1/3},\infty)$ and is explicitly given by
$$q_{min}(r)=\frac{1}{6}\left(\sqrt{(4r^3-3)(4r-3)}-4r^2\right).$$
Finally, studying the function 
$$h(r)=g(q_{min}(r),r)=-\left(\frac{\sqrt{(4r^3-3)(4r-3)}}{3}\right)^3+\frac{64r^6-72r^5+54r^4-72r^3+27}{27},$$
it can be shown by standard single-variable methods that $h(r)\geq0$ for $r\in(0,\frac34]\cup[(\frac34)^{1/3},\infty)$, and the minimal value $0$ is attained only for $r=0$ and $r=1$. $\blacksquare$
A: Let $\phi=\frac{2}{3}(-\pi-\theta)$, then the equation becomes 
$$|1+z\omega|^2|1+z\overline{\omega}|^2\geq\frac{1}{2}+\frac{1}{2}\cos 3\phi$$
whenever $z=re^{i\phi}$ and $\cos\phi\leq 1/2$.
Let $2\cos\phi=1-u$ and the equation becomes
$$4(1-r)^2(1+r+r^2+ru)+4r^2u^2\geq 3u^2-u^3$$
whenever $r>0$ and $u\in [0,3]$.
That looks manageable.
