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