Let $z_1,z_2$ be complex numbers with $|z_1|=|z_2|=1$. Prove that $|z_1 + 1| + |z_2 + 1| + |z_1z_2 + 1| \geq 2$

Question

Let $z_1,z_2$ be complex numbers with $|z_1|=|z_2|=1$. Prove that :- $$|z_1 + 1| + |z_2 + 1| + |z_1z_2 + 1| \geq 2$$

What I Tried:- From Triangle Inequality, we have :-

$$|z_1 + 1| + |z_2 + 1| + |z_1z_2 + 1| \geq |(1 + z_1)(1 + z_2) + 2|$$

From here, no other specific ideas came to my mind. I still did not use the fact, $|z_1| = |z_2| = 1$, so from Triangle Inequality again, we get :-

$$|z_1 + 1| + |z_2 + 1| \leq 4$$

The $2$ equations seem nicely similar, but I wasn't able to connect them nicely.

Can someone help me? Thank You.

Algebraic Solutions will be preferred over Geometric Solutions.

Answer 1

Let $z_1=\cos\alpha+i\sin\alpha$ and $z_2=\cos\beta+i\sin\beta$.

Thus, we need to probe that $$|\cos\frac{\alpha}{2}|+|\cos\frac{\beta}{2}|+|\cos\frac{\alpha+\beta}{2}|\geq1,$$ for which it's enough to prove that $$\cos\frac{\alpha}{2}+\cos\frac{\beta}{2}-\cos\frac{\alpha+\beta}{2}\geq1,$$ where $\alpha$ and $\beta$ in $\left[0,\frac{\pi}{2}\right]$ or $$\cos\frac{\alpha-\beta}{4}\geq\cos\frac{\alpha+\beta}{4},$$ which is obvious.