 Solution

Use the Minkowski metric with signature $(+---)$ and write $U=(U^0,\mathbf{u})$, $V=(V^0,\mathbf{v})$. Future-pointing timelike means $U^0>\Vert \mathbf{u}\Vert$ and $V^0>\Vert \mathbf{v}\Vert$. Thusso $U+V$ is again future-pointing timelike.

Let $a=\sqrt{g(U,U)}>0$ and $b=\sqrt{g(V,V)}>0$. In a Lorentz frame where $U=(a,0,0,0)$, the future-pointing vector $V$ has $V^0\ge b$. Consequently $g(U,V)=a{V}^0\ge ab$, with equality exactly when $V$ has no spatial component in that frame. HenceTaking positive square roots provesEquality holds exactly when $U,V$ are positive scalar multiples of each other. This is the reverse triangle inequality for future timelike vectors, not the Euclidean triangle inequality.