Does there exist a function $\psi:[0,\infty) \rightarrow \mathbb{R}$ such that......? Does there exist a function  $\psi:[0,\infty) \rightarrow \mathbb{R}$ such that $\psi (x) \geq 0$, $\psi'(x)\neq 0$, $\psi''(x) < 0$ and 
\begin{equation*}
\alpha\cdot \psi''(x)^2-\psi'(x)\,\psi'''(x)>0
\end{equation*}
with $0<\alpha <1$? 
 A: No. For a fixed $\alpha\in(0,1)$, the solution of the differential equation
$$ \alpha\cdot \psi''(x)^2 = \psi'(x)\psi'''(x) \tag{1}$$
is given by:
$$ \psi(x)=C_3+\frac{C_2}{2-\alpha}\left[(1-\alpha)x+C_1\right]^{1+\frac{1}{1-\alpha}}\tag{2}$$
where $C_1,C_2,C_3\geq 0$ in order to have that $\psi(x)$ is defined and positive over $\mathbb{R}^+$.
In such a way $\psi(x)$ is non-decreasing, too, over $\mathbb{R}^+$. However, $(2)$ gives:
$$ \psi''(x) = C_2\cdot \left[(1-\alpha)x+C_1\right]^{\frac{\alpha}{1-\alpha}}\geq 0\tag{3}$$
hence the answer to your question is negative.
A: It is not an answer but an observation : there is no solution if we take $\psi(x)=x^{1/\beta}$ for $x\geq0$ and $\beta>1$.
Inedeed, we have
$$\psi'(x)=\frac{1}{\beta}x^{1/\beta-1}\neq0,$$
$$\psi''(x)=\frac{1}{\beta}\left(\frac{1}{\beta}-1\right)x^{1/\beta-2}<0,$$
$$\psi'''(x)=\frac{1}{\beta}\left(\frac{1}{\beta}-1\right)\left(\frac{1}{\beta}-2\right)x^{1/\beta-3}$$
but if we ask for
$$\alpha\left(\psi''(x)\right)^2-\psi'(x)\psi'''(x)
=\frac{\alpha}{\beta^2}\left(\frac{1}{\beta}-1\right)^2x^{2/\beta-4}-\frac{1}{\beta^2}\left(\frac{1}{\beta}-1\right)\left(\frac{1}{\beta}-2\right)x^{2/\beta-4}>0$$
then
$$\alpha\left(\frac{1}{\beta}-1\right)^2-\left(\frac{1}{\beta}-1\right)\left(\frac{1}{\beta}-2\right)>0$$
i.e.
$$\alpha>\frac{\frac{1}{\beta}-2}{\frac{1}{\beta}-1}=\frac{2\beta-1}{\beta-1}>1.$$
