Integral inequality of length of curve 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.

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 $[a, b]$.
