integral inequality increasing continuous function. Let $f:[a,b]\rightarrow \mathbb R^+$ and $g:[a,b]\rightarrow \mathbb R^+$ be two non-decreasing continuous functions such that for all $x\in [a,b]$ we have $\int\limits_{a}^x \sqrt{f(t)}dt\leq \int\limits_{a}^x \sqrt{g(t)}dt$ with equality when $x=b$.
Prove that $\int\limits_{a}^b \sqrt{f(t)+1}dt\geq \int\limits_a^b \sqrt{g(t)+1}dt$
I think that we should somehow use the concavity of the square root function.
 A: Call $u(x)=\int_{a}^{x}\sqrt{f(t)}dt$ and $v(x)=\int_{0}^{x}\sqrt{g(t)}dt$
Then $\int_{a}^{x}\sqrt{f(t)+1}dt=\int_{a}^{x}\sqrt{(u'(t))^2+1}dt$ is the arc length of the graph of $u(t)$ between $a$ and $x$.
The given inequality tells us that the graph of $v$ is never below the graph of $u$. The two graphs meet at $x=a$ and at $x=b$.
But since $f,g$ are non-negative (and therefore $\sqrt{f},\sqrt{g}$ are non-negative), we have that the derivatives of $u$ and $v$ are always positive.
$f,g$ might not be differentiable, but the increment of the derivative of $u$ and $v$ can still be computed. For $s>0$
$$u'(t+s)-u'(t)=\sqrt{f(t+s)}-\sqrt{f(t)}=\frac{f(t+s)-f(t)}{\sqrt{f(t+s)}+\sqrt{f(t)}}\geq0$$
Therefore $u,v$ are convex, one below the other and meet at the two end-points. Therefore the one below has a longer arc than the one above.

The last step is still meaty, but since it is not too hard, well-known, and a bit tedious to explain, I left it out.
An argument, not completely precise, has been given for example in an answer to this previous question. 
