Prove the inequalities $|z_1 +z_2| \geq \frac{1}{2}(|z_1|+|z_2|)|\frac {z_1} {|z_1|} +\frac {z_2} {|z_2|}|$ Prove the inequalities $|z_1 +z_2| \geq \frac{1}{2}(|z_1|+|z_2|)|\frac {z_1} {|z_1|} +\frac {z_2} {|z_2|}|$
Here is what I got so far
I start with $z_1=r_1 cis \theta_1$ (at this time we are not allowed to use $e^{i\theta}$ but $cis \theta =e^{i \theta}$)
$z_2 =r_2 cis \theta_2$. I will focus on the right hand side of the inequality 
$RHS= \frac{1}{2}(r_1 +r_2)| \frac {r_1 cis \theta_1}{r_1} +\frac{r_2 cis \theta_2}{r_2}|=\frac{1}{2}(r_1 +r_2)|cis \theta_1 +cis \theta_2|=\frac{1}{2}(r_1 +r_2)((\cos \theta_1 +\cos \theta_2)^2 +(\sin \theta_1+\sin \theta_2)^2 )^{1/2}=\frac{1}{2}(r_1 +r_2)(2+2(cos \theta_1 cos \theta_2+ sin \theta_1 sin\theta_2))^{1/2}=\frac{1}{2}(r_1 +r_2)(2+2(cos (\theta_1 -\theta_2))^{1/2}=(r_1+r_2)(1+\cos(\theta_1-\theta_2))$
but I'm not sure how this result can help me prove the inequality, because I can't see how the left hand side related to this.
 A: The LHS becomes $\sqrt{r_1^2+r_2^2+2r_1r_2\cos(\theta_1-\theta_2)}$
As the LHS,RHS both are $\ge0,$
LHS will be $\ge RHS\iff$ LHS$^2\ge$RHS$^2$
Now, $r_1^2+r_2^2+2r_1r_2\cos(\theta_1-\theta_2)-\dfrac{(r_1+r_2)^2(1+\cos(\theta_1-\theta_2))}2$
$=\dfrac{(r_1-r)^2(1-\cos(\theta_1-\theta_2))}2$ which is definitely non-negative 
A: since
$$z_{1}=r_{1}(\cos{\theta_{1}}+ir_{1}\sin{\theta_{1}},z_{2}=r_{2}\cos{\theta_{2}}+ir_{2}\sin{\theta_{2}}$$
$$\Longrightarrow |z_{1}+z_{2}|=\sqrt{(r_{1}\cos{\theta_{1}+r_{2}\cos{\theta_{2}})^2+(r_{1}\sin{\theta_{1}}+r_{2}\sin{\theta_{2})^2}}}=\sqrt{(r^2_{1}+r^2_{2}+2r_{1}r_{2}\cos{(\theta_{1}-\theta_{2})})}$$
$$\Longleftrightarrow 2(r^2_{1}+r^2_{2}+2r_{1}r_{2}\cos{(\theta_{1}-\theta_{2})})\ge (r_{1}+r_{2})^2[1+\cos{(\theta_{1}-\theta_{2})}]$$
$$\Longleftrightarrow r^2_{1}+r^2_{2}+4r_{1}r_{2}\cos{(\theta_{1}-\theta_{2})}\ge 2r_{1}r_{2}+(r_{1}+r_{2})^2\cos{(\theta_{1}-\theta_{2})}$$
$$\Longleftrightarrow (r_{1}-r_{2})^2\ge (r_{1}-r_{2})^2\cos{(\theta_{1}-\theta_{2})}$$
It is clear
