On a sequence $(a_k)_{k\geq0}$, define the operator $\mathcal{L}a_k=a_ka_{k+2}-a_{k+1}^2$. Then, $(a_k)_k$ is called *log-convex* if $\mathcal{L}a_k\geq0$ for all $k\geq0$. One may also consider iteratively $\mathcal{L}^2a_k=\mathcal{L}\mathcal{L}a_k$ and $(a_k)_k$ is called *doubly log-convex* if $\mathcal{L}a_k\geq0$ as well as $\mathcal{L}^2a_k\geq0$ for all $k\geq0$.

I would like to ask:
>**QUESTION.** Suppose $(a_k)_k$ is doubly log-convex. Is the following true? If not, give a minimal added condition to secure it.
$$\frac{\mathcal{L}^2a_k\cdot\mathcal{L}^2a_{k+2}}{a_{k+2}\cdot a_{k+4}}
-\frac{(\mathcal{L}^2a_{k+1})^2}{(a_{k+3})^2}\geq0, \qquad \text{for all $k\geq0$}.$$