Triangle Inequality about complex numbers, special case At Spivak's calculus he treats the triangle inequality about complex numbers dividing into cases. Let $z$ and $w$ complex numbers, the inequality is clear if $z$ or $w$ are $0$. 
$\lvert z+w \rvert$$\leq \lvert z \rvert + \lvert w \rvert$
Suppose that $z=\lambda w$, then it is straightforward to prove the inequality but divide in cases :
$\lambda<0$  and $\lambda>0$. 
I can't see how is this straightforward and why do we need to divide into cases for $\lambda$.
After I do the algebraic manipulations it seems that it doesn't matter how $\lambda$ is.
My guess is that my approach is wrong, so if someone can give me a hint on how to initiate the proof would be a great help. Thank you
 A: A general approach : For real $a_1,a_2,b_1,b_2$: We have $$ \bullet \quad (a_1^2+b_1^2)(a_2^2+b_2^2)=(a_1b_2-a_2 b_1)^2+(a_1 a_2+b_1 b_2)^2\geq (a_1 a_2+b_1 b_2)^2.$$ 
For $z=a_1+i b_1$ and $w=a_2+i b_2$ we have $$|z+w|\leq |z|+|w|\iff |z+w|^2\leq (|z|+|w|)^2\iff $$  $$\iff  (a_1+a_2)^2+(b_1+b_2)^2\leq a_1^2+b_1^2+a_2^2+b_2^2 +2\sqrt {(a_1^2+b_1^2)(a_2^2+b_2^2)} \iff$$  $$ \iff 2 (a_1a_2+b_1b_2)\leq 2\sqrt {(a_1^2+b_1^2)(a_2^2+b^2)}.$$From $\bullet$ we have $$\sqrt {(a_1^2+b_1^2)(a_2^2+b_2^2)}\geq \sqrt {(a_1 a_2+b_1 b_2)^2}=|a_1a_2+b_1b_2|\geq a_1a_2+b_1b_2.$$ 
Remarks: The significance of the values $a_1b_2-a_2b_1$ and $a_1a_2+b_1b_2$ can be seen if we put $z$ and $w$ in polar form :$z=r_1(\cos t_1+i\sin t_1),\;w=r_2(\cos t_2+i\sin t_2)$ with real $t_1,t_2$ and non-negative real $r_1,r_2$. Then $a_1 b_2-a_2 b_1= r_1 r_2 \sin (t_2-t_1)$ and $a_1 a_2+b_1 b_2=r_1 r_2\cos (t_2-t_1).$ So  if $z\ne 0\ne w$  then $|z+w|=|z|+|w| \iff (\sin (t_2-t_1)=0\land \cos (t_2-t_1)=1) \iff z/w\in R^+.$ 
A: Let $z=(a_1,b_1), w=(a_2,b_2)$, suppose $z=lw$, then $z=(la_1,la_2)$
I want to prove $\lvert z+w \rvert \leq\lvert z\rvert+\lvert w\rvert$
calculate $\lvert z+w \rvert$ = $\bullet$:
$\bullet$=$\lvert (la_2,lb_2)+(a_2,b_2) \rvert$=$\lvert (la_2+a_2,lb_2+b_2)\rvert$= $\sqrt{(1+l)^2a^2+(1+l)^2b^2)}$=$\lvert 1+l \rvert$ $\sqrt{a^2+b^2}$
calculate $\lvert z \rvert+\lvert w \rvert$= $\star$ :
$\star$ =$\lvert (la_2,lb_2)\rvert$ + $\lvert (a_2,b_2) \rvert$= $\sqrt{l^2a^2+l^2b^2}$+$\sqrt{a^2+b^2}$=$\rvert l \lvert$ + 1 $\sqrt{a^2+b^2}$
if l>$0$ then $\lvert 1+l \rvert$=$\rvert l \lvert$ + 1 then $\lvert z+w \rvert =\lvert z\rvert+\lvert w\rvert$
if l
