How can I prove this inequality for complex numbers?

Question

Let be $a,b \in \mathbb{C}$. Prove that: $$|1+a|+|1+b|+|1+ab| \geq 2 \tag{1}$$

I prove it for $a,b\in \mathbb{R}$, but I cannot conclude the proof for complex. I prove that: $$|1+\Re(a)|^2 + |1+\Re(b)|^2+|1+\Re(ab)|^2 > |1+a|^2+|1+b|^2+|1+ab|^2 \tag{2}$$ Implies $|1+a|+|1+b|+|1+ab| \geq 2$, but I cannot conclude $(1)$ even suposing that: $$|1+\Re(a)| + |1+\Re(b)|+|1+\Re(ab)| > |1+a|+|1+b|+|1+ab| $$

Answer 1

If $|b|\ge 1$

$$\text{LHS}\ge|1+a|+|1+b-1-ab|=|1+a|+|b|\cdot|1-a|\ge|1+a|+|1-a|\ge2$$

If $|b|<1$ $$\text{LHS}\ge|b|\cdot|1+a|+|1+b|+|1+ab|\ge|1+b|+|b+ab-1-ab|=|1+b|+|1-b|\ge2$$