Inequality $|z_1+z_2|^2 \le (1+|z_1|^2)(1+|z_2|^2)$ I have a problem to prove this inequality 
$|z_1+z_2|^2 \le (1+|z_1|^2)(1+|z_2|^2)$ $\forall (z_1, z_2)\in \mathbb{C}$.
I tried to take the right hand set and subtract the lfs and after simplification I got this:
$1+(ax)^2+(by)^2 -2(ax+by)+(ay)^2+(bx)^2$  and I couldn't prove thqt this result is positive.
Any help please?

Put $z_k=r_k(\cos\theta_k+i\sin\theta_k),k=1,2$
$(z_1+z_2)^2=r_1^2+r_2^2+2r_1r_2\cos(\theta_1-\theta_2)$
$(1+|z_1|^2)(1+|z_2|^2)=(1+r_1^2)(1+r_2^2)=1+r_1^2+r_2^2+r_1^2r_2^2$ 
this will be $\ge r_1^2+r_2^2+2r_1r_2\cos(\theta_1-\theta_2)$
$\iff 1+r_1^2r_2^2\ge 2r_1r_2\cos(\theta_1-\theta_2)$
But $1+r_1^2r_2^2\ge 2r_1r_2$ using AM$\ge$ GM
and $r_1r_2\ge r_1r_2\cos(\theta_1-\theta_2)$ as  $\cos(\theta_1-\theta_2)\le1$
