Romanian Mathematical Olympiad

If there exists $k \in [-1, 1]$ such that $w = kv$, the inequality is obvious. Conversely, let $t > 1$ such that $w - tv \ne 0$. Setting
$$z = \frac{t\bar{v} - \bar{w}}{w - tv},$$
we have $|z| = 1$ and moreover
$$zw + \bar{w} = \frac{t(w\bar{v} - \bar{w}v)}{w - tv},$$
$$zv + \bar{v} = \frac{w\bar{v} - \bar{w}v}{w - tv},$$
implying
$$|zw + \bar{w}| = t |zv + \bar{v}|.$$
As $t > 1$, the given inequality yields $|zw + \bar{w}| = |zv + \bar{v}| = 0$, so $w\bar{v} - \bar{w}v = 0$, therefore $\frac{w}{v} = k \in \mathbb{R}$.
Plugging back in the initial condition we derive that $k \in [-1, 1]$, as needed.