TITLE: Integral inequality of length of curve
QUESTION [8 upvotes]: Let $f:\mathbb{R}\to \mathbb{R}$ be a continuously differentiable function. Prove that for any $a.b\in \mathbb{R}$
$$\left (\int_a^b\sqrt{1+(f'(x))^2}\,dx\right)^2\ge (a-b)^2+(f(b)-f(a))^2$$.

source: This is from TIFR GS stage 2.
I think mean value theorem kills it but can't do it ...even try Cauchy-Schwarz inequality but nothing conclusion.Geometrically its true since distance  between two point $(a,f(a))$ & $(b,f(b))$ is greater or equal to any arc length of same endpoints.

REPLY [15 votes]: 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.