I am looking at arc length of a continuous differentiable function $f$ in $\mathbb{R}^2$. Such that the arc length is $L = \int^{b}_{a}\sqrt{1+(\frac{dy}{dx})^2} dx$. Where the derivative of $f$ is $f' =\frac{dy}{dx}$.
When I think of the triangle inequality, the idea of the shortest distance between two points is the line in between them. So naturally I came up with $$|b-a| \leq \int^{b}_{a}\sqrt{1+(\frac{dy}{dx})^2} dx$$ Is this bound true? I would think if it is, you would most likely using the triangle inequality to prove it.
Edit: So after the answer from below here is the rough outline of the proof (so that it is all contained within the question)
$0 \leq x^2$
$0 \leq(\frac{dy}{dx})^2$
$1 \leq 1+(\frac{dy}{dx})^2$
$1 \leq \sqrt{1+(\frac{dy}{dx})^2}$
$\int^{b}_{a} 1 dx \leq \int^{b}_{a}\sqrt{1+(\frac{dy}{dx})^2} dx$
$ |b-a| \leq \int^{b}_{a}\sqrt{1+(\frac{dy}{dx})^2} dx$
The arc length is a non-negative real number (and the distance between two points cannot be negative), so the above is indeed true for $b\geq a$.
Since $(\frac{dy}{dx})^2\geq 0$ we have that $$ \int^{b}_{a}\sqrt{1+(\frac{dy}{dx})^2} dx\geq\int_{a}^{b}1dx=b-a.$$