Proving this equality by using module properties

Question

I Want to prove that

$2(|z| + |w|) = |z+w+2 \sqrt(zw)| + |z+w-2 \sqrt(zw)|$

my ideia was to use module properties and show that

$2(|z| + |w|) \leq |z+w+2 \sqrt(zw)| + |z+w-2 \sqrt(zw)|$

and

$2(|z| + |w|) \geq |z+w+2 \sqrt(zw)| + |z+w-2 \sqrt(zw)|$

but "breaking" the modules above and manipulating them i always get a mix of $\leq$ and $\geq$ in one inequalitie that get me to nowhere.

Answer 1

$(|z +w + 2\sqrt{zw}| + |z + w - 2\sqrt{zw}|)^2 \\= 2|z|^2 + 2|w|^2 + 4<z,w> + 8 |\sqrt{zw}|^2 + 2|(z+w+2\sqrt{zw})(z+w-2\sqrt{zw})|\\ = 4|z|^2 + 4|w|^2 + 8 |(\sqrt{zw})^2|\\ = 4(|z| + |w|)^2$

Then take the square root of both sides. The notation $<\cdot,\cdot>$ means the scalar product in $\mathbb{C}$.