Arc Length of a Curve Let $f:[a,b]\to \mathbb{R}$ be a continuous function, how can you prove (not in the geometric way):
$$
\sqrt{\left(f(b)-f(a)\right)^2+\left(b-a\right)^2}\le\int_a^b \sqrt{1+f'(x)^2}dx
$$
 A: 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
$$\sqrt{\sum_{i=1}^n\left(f_i(b)-f_i(a)\right)^2} \leq \int_a^b\sqrt{\sum_{i=1}^n f_i'(x)^2}dx$$
A: Hint: $$\sqrt{\left(f(b)-f(a)\right)^2+\left(b-a\right)^2} = (b-a)\sqrt{\left(\frac{f(b)-f(a)}{b-a}\right)^2+1} $$
A: More generally you can proof that $\sqrt {\left( \int_a^b f\right)^2+\left( \int_a^b g\right)^2}\leq \int_a^b\sqrt{f^2+g^2}$ differentiating by b.
In fact you obtaine $$\frac{f\int_a^b f+g\int_a^b g}{\sqrt{\left(\int_a^b f \right)^2+\left(\int_a^b g\right)^2}}\stackrel{?}{\leq} \sqrt{f^2+g^2}$$ 
that is $0\stackrel{?}{\leq} \left( f \int_a^b g- g \int_a^bf \right)^2$.
A: Hint
The lhs is the distance between points $[a,f(a)]$ and $[b,f(b)]$ along the line which joins them. And since the shortest distance between two points is ...
I am sure that you can take from here.
