Triangle inequality- complex I am trying to prove the triangle inequality purely algebraically. 
Let $z=x+iy$, $w=u+iv$. 
Then, 
$|z+w|^2$=$|(x+u)+i(y+v)|^2$=$(x+u)^2+(y+v)^2$=$x^2+2xu+u^2+y^2+2yv+v^2$
I tried the other way:
$(|z|+|w|)^2$=$(\sqrt{x^2+y^2}+\sqrt{u^2+v^2})^2$=$x^2+y^2+u^2+v^2+2 \sqrt{x^2+y^2} \sqrt{u^2+v^2}$
I'm trying to do this without using the complex conjugate or anything. 
Any help is appreciated. 

Hint: Use Cauchy-Schwarz inequality:
$$
\sqrt{x^2+y^2} \sqrt{u^2+v^2}\geq xu+yv
$$

To prove it directly:
$$
\sqrt{x^2+y^2} \sqrt{u^2+v^2}\geq xu+yv\implies\\
x^2u^2+x^2v^2+y^2u^2+y^2v^2\geq x^2u^2+2xvyu+y^2v^2\implies\\
x^2v^2+y^2u^2 - 2xvyu\geq 0 \implies\\
(xv-yu)^2 \geq 0 \\
$$
