inequality with modulus of complex number Let $ \displaystyle{ z_1, z_2 \in \mathbb{C} }$ where $ z_1, z_2 \neq 0$
Prove that: $\displaystyle |z_1 +z_2| \geq \frac{1}{2} \left( |z_1|+|z_2| \right) \left|\frac{z_1}{|z_1|} + \frac{z_2}{|z_2|}\right| $.
P.S I think that I have to use the inequality $ Re(z_1z_2) \leq |z_1||z_2| $ but I don't know how.
 A: Write $z_1=r_1e^{i\theta_1}$ and $z_2=r_2e^{i\theta_2}$. Since $z_1, z_2\neq 0$, we have $r_1, r_2>0$. Then
$$\tag{1}\left(\frac{1}{2} \left( |z_1|+|z_2| \right) \left|\frac{z_1}{|z_1|} + \frac{z_2}{|z_2|}\right|\right)^2
=\left(\frac{1}{2}(r_1+r_2)|e^{i\theta_1}+e^{i\theta_2}|\right)^2=\frac{1}{4}(r_1+r_2)^2|e^{i\theta_1}+e^{i\theta_2}|^2$$
$$=\frac{1}{4}(r_1+r_2)^2(2+e^{i(\theta_1-\theta_2)}+e^{i(\theta_2-\theta_1)})$$
since $$\tag{2} |e^{i\theta_1}+e^{i\theta_2}|^2=(e^{i\theta_1}+e^{i\theta_2})(e^{-i\theta_1}+e^{-i\theta_2})=2+e^{i(\theta_1-\theta_2)}+e^{i(\theta_2-\theta_1)}.$$ Note also that 
$$\tag{3}|z_1+z_2|^2=|r_1e^{i\theta_1}+r_2e^{i\theta_2}|^2=
r_1^2+r_2^2+r_1r_2e^{i(\theta_1-\theta_2)}+r_1r_2e^{i(\theta_2-\theta_1)}.$$
Subtract $(3)$ by $(1)$, we obtain
$$|z_1+z_2|^2-\left(\frac{1}{2} \left( |z_1|+|z_2| \right) \left|\frac{z_1}{|z_1|} + \frac{z_2}{|z_2|}\right|\right)^2=\frac{1}{2}(r_1-r_2)^2-\frac{1}{4}(r_1-r_2)^2\big(e^{i(\theta_1-\theta_2)}+e^{i(\theta_2-\theta_1)}\big)$$
$$=\frac{1}{4}(r_1-r_2)^2(2-e^{i(\theta_1-\theta_2)}-e^{i(\theta_2-\theta_1)})
=\frac{1}{4}(r_1-r_2)^2|e^{i\theta_1}-e^{i\theta_2}|\geq 0.$$
where the last equality follows from a caluculation similar to $(2)$. So this implies that
$$\displaystyle |z_1 +z_2| \geq \frac{1}{2} \left( |z_1|+|z_2| \right) \left|\frac{z_1}{|z_1|} + \frac{z_2}{|z_2|}\right|.$$
