Let $h,k:[a,b]\to \mathbb R$ be two $C^1$ functions which coincides at the extremes and such that $h\leq k$ on $[a,b]$ and let the graph of $h$ be an arc of a circle of radius $R$ curved upwards. Show that $$L(k)-L(h)\geq \frac 1R|D|,$$ Where $L(k)$ and $L(h)$ denotes the lengths of the graph of $k$ and $h$ respectively, and $|D|$ is the area of the region $D$ between the two graphs.
My attempt: Playing around and recalling that the length of a graph on an interval is related to the quantity $$\sqrt{1+[f'(x)]^2},$$ I've studied the following inequality $$\sqrt{1+y^2}-\sqrt{1+x^2},$$ and it seems to me that, using the mean value theorem, and the fact that $$\left(\frac{x}{\sqrt{1+x^2}}\right)^'=\left(\frac{1}{(1+x^2)^{3/2}}\right)>0$$ that the following inequality holds, $\forall x,y\in\mathbb R$:
$$\sqrt{1+y^2}-\sqrt{1+x^2}\geq \frac{x}{\sqrt{1+x^2}}(y-x).$$ Please correct me if I am wrong. My plan now is to use this inequality to recover information about the inequality of the problem, but i cannot go further.
Any help is appreciated. Thank you very much.
Edit: I admit your objection is valid, thanks @Beni, i add the hypothesis. And now? How to prove the statement which now should be well posed I think.