If $\gamma : [a,b] \to \Bbb{R}^n$ continuously differentiable, show that the arc length $arc(\gamma)$ equals $\sup \{ \sum_{i=1}^n ||\gamma (t_i)-\gamma (t_{i-1})|| \mid a = t_0 < t_1 < ... < t_n = b \}$.
Recall that $arc(\gamma ) = \int_{a}^{b} ||\gamma '(t)||dt$. I'm trying to solve the above problem. I've already shown that a sequence of points in $\{ \sum_{i=1}^n ||\gamma (t_i)-\gamma (t_{i-1})|| \mid a = t_0 < t_1 < ... < t_n = b \}$ converges to $arc (\gamma)$, so I just need to show that $arc (\gamma)$ is an upper bound of the set. I've reduced this to showing that $\int_{a}^{b} ||\gamma ' (t)|| dt \ge ||\gamma (b) - \gamma (a)||$ holds. Intuitively/geometrically it's obvious, but I having trouble mathematizing my intuition. I could use some help.
You have essentially solved this already. $$ |\gamma(b)-\gamma(a)| = \left| \int_a^b \gamma'(t) \, dt \right| \le \int_a^b |\gamma'(t)| \, dt $$
by the triangle inequality. This is another version of the Mean Value Theorem, summarisable as: "global growth $\le$ local growth".