Reverse triangle inequality (proper-time maximization)

**Step 1 (adapt the inertial frame to $\gamma_0$).** Because $\gamma_0$ is a straight timelike line, we can choose an inertial coordinate system $(t,x,y,z)$ in which $\gamma_0$ sits at rest at the spatial origin: $p=(0,\mathbf 0)$, $q=(T,\mathbf 0)$ with $T>0$, and $\gamma_0(t)=(t,\mathbf 0)$ for $t\in[0,T]$. Since the metric $ds^2 = -c^2 dt^2 + dx^2 + dy^2 + dz^2$ is invariant under Lorentz transformations, the proper time of any curve is the same in every inertial frame.

**Step 2 (compute $\tau(\gamma_0)$).** Along $\gamma_0$ the spatial velocity is $\mathbf v(t)=\mathbf 0$, so $$\tau(\gamma_0) = \int_0^T \sqrt{1 - 0/c^2}\,dt = T.$$

**Step 3 (parameterize the competing curve $\gamma$).** Any future-directed timelike curve $\gamma$ from $p$ to $q$ has $dt/d\lambda > 0$ everywhere (since $c^2 dt^2 > |d\mathbf x|^2 \ge 0$), so we may parameterize it by the coordinate time $t\in[0,T]$ as $\gamma(t)=(t,\mathbf x(t))$ with $\mathbf x(0)=\mathbf x(T)=\mathbf 0$ and velocity $\mathbf v(t)=d\mathbf x/dt$ satisfying $|\mathbf v(t)|<c$. Its elapsed proper time is $\tau = \int \sqrt{1 - v(t)^2/c^2}\,dt$ over $[0,T]$.

Step 4 (pointwise bound and equality case). For every $t\in[0,T]$, $|\mathbf v(t)|^2\ge0$ implies $0 < \sqrt{1 - |\mathbf v(t)|^2/c^2} \le 1$, with equality at a given $t$ if and only if $\mathbf v(t)=\mathbf 0$. Integrating over $[0,T]$, $$\tau(\gamma) = \int_0^T \sqrt{1 - \frac{|\mathbf v(t)|^2}{c^2}}\,dt \le \int_0^T 1\,dt = T = \tau(\gamma_0).$$ Because the integrand is continuous and $\le 1$, equality $\tau(\gamma)=T$ holds if and only if $\mathbf v(t)=\mathbf 0$ for all $t\in[0,T]$, which together with $\mathbf x(0)=\mathbf 0$ forces $\mathbf x(t)\equiv\mathbf 0$, i.e. $\gamma=\gamma_0$.