Verify that $\sqrt{2}|z| \geq | R_z|+|Im_z|$

Question

Verify that $\sqrt{2}|z| \geq | R_z|+|Im_z|$, suggestion: Reduce this inequality to $(|x|-|y|)^2 \geq0$ (z is a complex number. R stands for real part and Im stands for imaginary part)

Approach: Let $z=x+yi$ where $x,y \in R$. By the triangle property, we have the following $$|x+yi|\geq |x|+|yi|$$

If we multiply the left side by $\sqrt{2}$, the inequality is preserved but who knows if there is equality. That's my problem

Answer 1

Let $z=a+bi$, $a,b \in \mathbb{R}$. Note $$a^2+b^2 \geq 2ab.$$ This follows immediately from the fact that $(a-b)^2 \geq 0$. So $$2(a^2+b^2) \geq a^2+b^2+2ab.$$ Now we take square roots at both sides to get the desired result.

Answer 2

Square both sides to get

$$2|z|^2=2(R_z^2+I_z^2)\ge R_z^2+I_z^2+2|R_z||I_z|$$

Subtracting the right-hand side from the left-hand side gives

$$R_z^2+I_z^2-2|R_z||I_z|=(|R_z|-|I_z|)^2\ge 0$$