Prove that for $\displaystyle z \in \mathbb{C}$, $\displaystyle |z| \leq |Re \ z| + |Im \ z| \leq \sqrt{2}|z|$.
Surely you that $\displaystyle |a|\le |a|+ |b|$. You also need to know that $\displaystyle 2|a||b|\le |a|^2+ |b|^2 $.
Look at $\displaystyle \left( {\left| a \right| + \left| b \right|} \right)^2 \leqslant \left| a \right|^2 + 2\left| a \right|\left| b \right| + \left| b \right|^2 \leqslant 2\left| a \right|^2 + 2\left| b \right|^2 $
Taking the square root, $\displaystyle \left( {\left| a \right| + \left| b \right|} \right) \leqslant \sqrt 2 \sqrt {\left| a \right|^2 + \left| b \right|^2 } $
Now let $\displaystyle a=\text{Re}(z)~\&~ b=\text{Im}(z).$