Let $|a| + |b| \le 1$ and let $z \in \mathbb{C}$ with $|z| = 1$. Then $|az + b\bar{z}| \le |az| + |b\bar{z}| = |a| + |b| \le 1$, as claimed.

Conversely, if $a = 0$ or $b = 0$ there is nothing to show. For $a, b \neq 0$, write $\frac{b}{a} = r(\cos\alpha + i\sin\alpha)$. Put $z = \cos\frac{\alpha}{2} + i\sin\frac{\alpha}{2}$ to get $1 \ge |az + b\bar{z}| = |a||\bar{z}|\left|z^2 + \frac{b}{a}\right| = |a|(1+r)(\cos\alpha + i\sin\alpha) = |a|(1+r) = |a|(1 + \left|\frac{b}{a}\right|) = |a| + |b|$, which ends the proof.