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.