Suppose that $$\int_a^{b} \|\alpha'(t)\|dt=\|\alpha(b)-\alpha(a)\|,$$ where $\alpha$ is a regular curve of class $C^1$. Prove that alpha's trace is the segment $\overline{AB}$, where $A=\alpha(a)$, and $B=\alpha(b)$.
Attempt at a solution: Mutiplying the first equation with $\|\alpha(b)-\alpha(a)\|$, we get $$\int_a^b\|\alpha'(t)\|\cdot \|\alpha(b)-\alpha(a)\|dt=\int_a^b \langle \alpha'(t),\alpha(b)-\alpha(a) \rangle dt.$$ I tried working with Schwarz's inequality, but couldn't get anywhere. I have already proven that $$\|\alpha(b)-\alpha(a)\| \leq \int_a^b \|\alpha'(t)\|dt$$ though.
You are given $$\|\alpha(b)-\alpha(a)\| = \int_a^b \|\alpha'(t)\|\,dt.$$ Let's write $c = \alpha(b)-\alpha(a)$ to save typing and improve legibility. Also, by the Fundamental Theorem of Calculus, $$\int_a^b \alpha'(t)\,dt = \alpha(b)-\alpha(a) = c.$$ Therefore, $$\|c\|^2=\left\langle\int_a^b \alpha'(t)\,dt,c\right\rangle = \int_a^b \langle \alpha'(t),c\rangle\,dt.$$ But $$ \int_a^b \langle \alpha'(t),c\rangle\,dt \le \int_a^b \|\alpha'(t)\|\|c\|\,dt = \|c\|^2.$$ Now, equality holds if and only if $\langle\alpha'(t),c\rangle = \|\alpha'(t)\|\|c\|$ for each $t\in [a,b]$ (you need a continuity argument to prove this) if and only if $\alpha'(t)$ is a positive scalar multiple of $c$ for each $t\in [a,b]$. And we are given that equality holds. Therefore, we conclude that $\alpha'(t)$ is a positive scalar multiple of $c$, which tells us that the curve $\alpha(t)$ is a portion of a straight line, as required.