Is this function involving square root smooth? Let $\psi:[0,1] \to \mathbb{R}$ be a concave, smooth, strictly increasing function satisfying $\psi(0) = 0$, $\psi(1) = 1$ and $\psi'(0)>1$. Assume further that $\psi$ is linear in a neighbourhood of zero, and set $c = 2\psi'(0)$. 
Note that the assumptions $\psi'(0)>1,\psi(1)=1$ imply that $\psi$ cannot be linear up to $r=1$-it must become strictly concave at some point.
Set $t_0=\sup\{ \psi'(r)+\frac{\psi(r)}{r}=c\}$. 

Question: Is $f(r)=\sqrt{c^2-(\psi'(r)+\frac{\psi(r)}{r})^2}$ infinitely differentiable at $t_0$?

As I explain below, $f(r)=0$ for every $r\le t_0$. So, this is equivalent to asking whether all the right the derivatives of $f(r)$ exist and are equal to zero at $t_0$.
Here are the details:
First, we note that the function $g(r)= \psi'(r)+\frac{\psi(r)}{r}$ is non-increasing, due to the concavity of $\psi$ (see a proof at the end). 
Also, $\lim_{r \to 0}g(r)=2\psi'(0)=c$. These facts implies that $g(r) \le c $ for every $r>0$, and that $g(r)=c$ on $[0,t_0]$. Equivalently, $\psi|_{[0,t_0]}$ is the solution to the ODE $y(r)'+y(r)/r=c$ which implies that $\psi(r)$ is linear on $[0,t_0]$.
The fact that $g(r)$ is non-increasing implies that $g(r)<c$ for every $r>t_0$.

As explained in this partial answer, the smoothness of $\psi$ implies that $$\sqrt{c^2 - \left(\psi'(t_0+h) + \frac{\psi(t_0+h)}{t_0+h}\right)^2} = o(h^n),$$
for any $n>1$. However, unfortunately, this fact alone does not imply that this creature is smooth at $t=t_0$.

A proof that $g(r)$ is non-increasing:
$$
g'(r)=\psi''(r)+\frac{1}{r}(\psi'(r)-\frac{\psi(r)}{r}),
$$
and both summands are non-positive. $\psi'' \le 0$ by concavity. Since
$\psi(r)=\int_0^r \psi'(t)dt \ge \int_0^r \psi'(r)dt=r\psi'(r)$, the second summand is also non-positive.
 A: Since $\psi$ is smooth and is linear up to $t_0$, you have that
$$\psi(t_0) = ct_0/2, \quad \psi'(t_0) = c/2, \quad \psi^{(n)}(t_0) = 0$$ for $n>1$.
This gives you, for any $n>1$,
$$\psi(t_0+h) = \frac{c}{2}(t_0+h) + o(h^n), \qquad \psi'(t_0+h) = \frac{c}{2} + o(h^n).$$
Therefore,
$$c^2 - \left(\psi'(t_0+h) + \frac{\psi(t_0+h)}{t_0+h}\right)^2 = o(h^n),$$
for any $n>1$.
In particular, you obtain that $f(r)$ has a zero derivative at $r=t_0$.
Let us now analyze $f'(r)$ as $r\to t_0^+$.
For simplicity, let us define $\eta(r) = \psi(r) - \frac{c}{2}r$, and $h(r) =-\eta'(r) - \frac{\eta(r)}{r}$.
Note that $h$ and all its derivatives vanish at $r=t_0$, and that $h$ is non-negative.
We then have, for $r>t_0$,
$$f(r) = \sqrt{2c\,h(r) + h^2(r)},$$
hence
$$f'(r) = \frac{ch' + hh'}{\sqrt{2c h + h^2}}= \frac{h'}{\sqrt{h}}\frac{c + h}{\sqrt{2c + h}}.$$
We therefore obtain that $f'(r)\to 0$ as $r\to t_0^+$; indeed, this follows from
$$\lim_{r\to t_0^+} \frac{(h'(r))^2}{h(r)} = \lim_{r\to t_0^+} \frac{2h'(r)h''(r)}{h'(r)} = \lim_{r\to t_0^+} 2h''(r) = 0.$$
Thus we obtained that $f(r)$ is $C^1$.
I expect that you can continue similarly for higher derivatives.
