Integral inequality of length of curve

Notice that the function $y \mapsto \sqrt{1+y^2}$ is strictly convex. So by the Jensen's inequality,

$$ \frac{1}{b-a} \int_{a}^{b} \sqrt{1 + f'(x)^2} \, \mathrm{d}x \geq \sqrt{1 + \left(\frac{1}{b-a}\int_{a}^{b} f'(x) \, \mathrm{d}x\right)^2} = \sqrt{1 + \left(\frac{f(b) - f(a)}{b-a} \right)^2}. $$

Multiplying both sides by $b-a$ and squaring proves the desired inequality. Moreover, by the strict convexity, the equality holds if and only if $f'$ is constant over $[a, b]$.

Note that for every complex valued integrable function $\phi :[a,b]\to \Bbb C$, it holds that $$ \left|\int_a^b \phi(x)\ dx\right|\le \int_a^b|\phi(x)|\ dx. $$ Let $\phi(x)=1+if'(x)$. Then we can see that $$\begin{align*} \left|\int_a^b \phi(x)\ dx\right|&=\left|(b-a)+i(f(b)-f(a))\right|\\&=\sqrt{(b-a)^2+(f(b)-f(a))^2} \end{align*}$$ and $$ \int_a^b|\phi(x)|\ dx=\int_a^b \sqrt{1+(f'(x))^2}\ dx. $$ Now, the desired inequality follows.
Note: The equality holds when $\text{arg}(\phi(x))$ is constant, that is, $\frac{f'(x)}{1}=f'(x)$ is constant.

An easy way to do this is to note that since distance is invariant under rotations, without loss of generality, we may assume that $f(a)=f(b).$ And now, since $\sqrt{1-f'(x)}\ge 0$ on $[a,b]$, the function in $C^1([a,b])$ that minimizes the integral coincides with the function $f$ that minimizes the integrand, and clearly, this happens when $f'(x)=0$ for all $x\in [a,b].$ That is, when $f$ is constant on $[a,b].$ Then, $f(x)=f(a)$ and the result follows.

If you want to do this without the wlog assumption, then argue as follows:

Let $\epsilon>0,\ f\in C^1([a,b])$ and choose a partition $P=\{a,x_1,\cdots,x_{n-2},b\}$.

The length of the polygonal path obtained by joining the points

$(x_i,f(x_i))$ is $\sum_i \sqrt{(\Delta x_i)^2+(\Delta y_i)^2}$ and this is clearly $\ge (b-a)^2+(f(b)-f(a))^2$. (You can make this precise by using an induction argument on $n$.)

And this is true for $\textit{any}$ partition $P$.

But the above sum is also $\sum_i\sqrt{1+\frac{\Delta y_i}{\Delta x_i}}\Delta x_i $ and now, upon applying the MVT, we see that what we have is a Riemann sum for $\sqrt{1+f'(x)}$.

To finish, choose $P$ such that $\left |\int^b_a\sqrt{1+f'(x)}dx- \sum_i\sqrt{1+f'(c_i)}\Delta x_i \right |<\epsilon $. (The $c_i$ are the numbers $x_i<c_i<x_{i-1}$ obtained from the MVT). Then,

$(b-a)^2+(f(b)-f(a))^2\le \sum_i\sqrt{1+f'(c)}\Delta x_i<\int^b_a\sqrt{1+f'(x)}+\epsilon.$

Since $\epsilon$ is arbitrary, the result follows.

For a slick way to do this, use a variational argument: assuming a minimum $f$ exists, consider $f+t\phi$ where $t$ is a real parameter and $\phi$ is arbitrary $C^1([a,b])$.

Subsitute it into the integral:

$l(t)=\int_a^b \sqrt{1+(f'+t\phi')^2}dx$.

Since $f$ minimizes this integral, the derivative of $l$ at $t=0$ must be equal to zero. Then,

$0=l'(0)= \int_a^b \dfrac{f'\phi'}{\sqrt{1+(f')^2}}dx$.

After an integration by parts, we get

$\dfrac{f'}{\sqrt{1+(f')^2}} = c$ for some constant $c\in \mathbb R,$ from which it follows that $f'=c$. And this means, of course, that the graph of $f$ is a straight line connecting $(a,f(a))$ and $(b,f(b)).$ The desired inequality follows.