Show the following equality Basically I want to show the following:
$$\sqrt{2}\ |z|\geq\ |\operatorname{Re}z| + |\operatorname{Im}z|$$
So what I did is the following:
Let $z = a + bi$
Consider the following:
$$2|z|^2 = 2a^2 + 2b^2 = a^2 + b^2 +a^2+b^2$$
Since $(a-b)^2\geq0$, hence $a^2+b^2\geq 2ab$
Thus $2|z|^2 \geq a^2+b^2+2ab = (a+b)^2$
Hence $\sqrt{2}|z| \geq a + b$ but $a = |\operatorname{Re}z|~,~ b = |\operatorname{Im}z|$
Did I make a mistake somewhere, if yes I would appreciate it if it could be pointed out and perhaps provide some guideline on how to prove this.
 A: Let $z = r\cos\theta + ri\sin\theta$
Then $|Re(z)| + |Im (z)| = \pm\sqrt 2 r\sin(\theta \pm \frac{\pi}{4}) \leq \sqrt 2 r$
All four possible sign combinations will cover $\theta$ in all quadrants. 
A: Let $z=a+bi$. Then $|z|^2=a^2+b^2$ and Re$z=a$ and Im$z=b$. Observe that
$$(\sqrt{2}|z|)^2=2(a^2+b^2)$$ and $$\Big(|\text{Re}z|+|\text{Im}z|\Big)^2=a^2+2|ab|+b^2.$$
Now,
$$
\begin{align}
(|a|-|b|)^2\geq 0&\iff a^2+b^2-2|ab|\geq 0\\
&\iff2(a^2+b^2)\geq a^2+2|ab|+b^2\\
&\iff (\sqrt{2}|z|)^2\geq\Big(|\text{Re}z|+|\text{Im}z|\Big)^2\end{align}
$$
Since $\sqrt{2}|z|$ and $|\text{Re}z|+|\text{Im}z|$ are nonnegative, the result follows.
A: Observe
\begin{align}
|a|\cdot 1+|b|\cdot 1 \leq \sqrt{a^2+b^2}\sqrt{1^2+1^2}.
\end{align}
A: It suffices to consider $x, y > 0$. Other case simply changes the sign of $x,y$ and of $\theta$. Thus:  $\dfrac{|\text{Re(z)}|}{|z|} = \dfrac{x}{\sqrt{x^2+y^2}}= \sin \theta, \dfrac{\text{Im(z)}}{|z|}= \dfrac{y}{\sqrt{x^2+y^2}} = \cos \theta, \theta \in \left(0,\dfrac{\pi}{2}\right)$. Thus you prove: $\sin \theta + \cos \theta \le \sqrt{2}$, which is true since $\sin \theta + \cos \theta = \sqrt{2}\sin\left(\theta+\dfrac{\pi}{4}\right)\le \sqrt{2}$ .
