<s>[INST] 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.
 [/INST] 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 $\sin \theta + \cos \theta = \sqrt{2}\sin\left(\theta+\dfrac{\pi}{4}\right)\le \sqrt{2}$ .
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