<s>[INST] Proving an inequality concerning arbitary complex numbers $\def\abs#1{\left|#1\right|}$If $y$ and $z$ are any complex numbers then prove that
\[
 2 \abs{y+z}\ge \bigl(\abs y + \abs z\bigr) 
\abs{\frac y{\abs y} + \frac z{\abs z}} \]
 [/INST] $\def\abs#1{\left|#1\right|}$If $y=r(\cos A+i\sin A), z=R(\cos B+i\sin B)$
$|y+z|=\sqrt{r^2+R^2+2Rr\cos(A-B)}$
$$ \abs{\frac y{\abs y} + \frac z{\abs z}}=\sqrt{2+2\cos(A-B)}$$
$2|y+z|$ will be $\ge (|y|+|z|)(\abs{\frac y{\abs y} + \frac z{\abs z}}) $
if $2\sqrt{r^2+R^2+2Rr\cos(A-B)} \ge (r+R)\sqrt{2+2\cos(A-B)}$
if $4(r^2+R^2+2Rr\cos(A-B)) \ge (r+R)^2(2+2\cos(A-B))$ 
if $2(1-\cos(A-B))(R-r)^2\ge 0$ which is true
</s>