TITLE: Arc Length of a Curve
QUESTION [6 upvotes]: Let $f:[a,b]\to \mathbb{R}$ be a continuous function, how can you prove (not in the geometric way):
$$
\sqrt{\left(f(b)-f(a)\right)^2+\left(b-a\right)^2}\le\int_a^b \sqrt{1+f'(x)^2}dx
$$

REPLY [8 votes]: Apply Cauchy–Schwarz on the vectors $u=(f'(x),1)$ and $v=(f(b)-f(a),b-a)$ to find
$$f'(x)(f(b)-f(a)) + (b-a) \leq \sqrt{f'(x)^2 + 1} \sqrt{\left(f(b)-f(a)\right)^2 + \left(b-a\right)^2}$$
Now integrate the inequality over $[a,b]$ to obtain the result
$$\sqrt{\left(f(b)-f(a)\right)^2 + \left(b-a\right)^2} \leq \int_a^b\sqrt{f'(x)^2 + 1}dx$$
${\bf Addendum}$: By considering $u=(f_1'(x),f_2'(x),\ldots,f_n'(x))$ and $v=(f_1(b)-f_1(a),\ldots,f_n(b)-f_n(a))$ the result easily generalizes to
$$\sqrt{\sum_{i=1}^n\left(f_i(b)-f_i(a)\right)^2} \leq \int_a^b\sqrt{\sum_{i=1}^n f_i'(x)^2}dx$$