# Proving an inequality concerning arbitary complex numbers

$\def\abs#1{\left|#1\right|}$If $y$ and $z$ are any complex numbers then prove that $2 \abs{y+z}\ge \bigl(\abs y + \abs z\bigr) \abs{\frac y{\abs y} + \frac z{\abs z}}$

-
Perhaps one of the pluses on the right-hand side should have been a multiplication? –  Henning Makholm Oct 11 '12 at 8:44
@Henning Makholmes:- Yes , multiplication it is. –  Souvik Dey Oct 11 '12 at 8:56

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

-
You mean y=r(cosA+isinA) , z=R(cosB+isinB) ? –  Souvik Dey Oct 11 '12 at 9:00
@Souvik, sorry for the missing parenthesis, rectified. –  lab bhattacharjee Oct 11 '12 at 9:01

The inequality is not true in general. Let $y=1, z=-1$ we have $$2|y+z|=0<2=(|y|+|z|)+\left|\frac{y}{|y|}+\frac{z}{|z|}\right|$$

-