Verify that $\sqrt{2}\mid{z}\mid$ is greater than or equal to $\mid$Re(z) $\mid$ + $\mid$Im(z) $\mid$ for any complex number z $\in$ C
2. This true for all real numbers: $a^2+b^2\ge 2ab$.
So $2(a^2+b^2)=a^2+(a^2+b^2)+b^2\ge a^2 +2ab + b^2=(a+b)^2$
Let $a=|\Re(z)|~\&~b=|\Im(z)|$ then does $2|z|^2\ge (a+b)^2$ follow?