Note that all members of the sequence are positive real numbers. The recursive relation gives
$$2x_n^2 = n + 2x_{n-1}x_{n-2} \le x_{n-1}^2 + x_{n-2}^2 + n.$$

$$\begin{aligned} 2x_n^2 &\le x_{n-1}^2 + x_{n-2}^2 + n, \\ 2x_{n-1}^2 &\le x_{n-2}^2 + x_{n-3}^2 + n-1, \\ &\vdots \\ 2x_3^2 &\le x_2^2 + x_1^2 + 3, \\ 2x_2^2 &\le x_1^2 + x_0^2 + 2. \end{aligned}$$
Adding all these inequalities gives
$$2x_n^2 + x_{n-1}^2 \le 2x_1^2 + x_0^2 + [n + (n-1) + \dots + 3 + 2],$$
i.e.
$$2x_n^2 + x_{n-1}^2 \le \frac{n^2 + n + 4}{2}. \quad (*)$$
If there exists a sought number $A$, the left hand side of this inequality is at least $2A^2n^2 + A^2(n-1)^2$, thus
$$2A^2n^2 + A^2(n-1)^2 \le \frac{n^2 + n + 4}{2},$$
i.e. $6A^2n^2 - 4A^2n + 2A^2 \le n^2 + n + 4$. This inequality holds for every $n \in \mathbb{N}$ only if $6A^2 \le 1$, i.e. $A \le \frac{\sqrt{6}}{6}$.
We prove that $A = \frac{\sqrt{6}}{6}$ satisfies the conditions of the problem.
First, we show inductively that $x_n > \frac{n\sqrt{6}}{6}$, for all $n \in \mathbb{N}_0$. Statement is true for $n=0$ and $n=1$. Let us assume $x_{n-1} > \frac{(n-1)\sqrt{6}}{6}$ and $x_{n-2} > \frac{(n-2)\sqrt{6}}{6}$. Then
$$x_n^2 = \frac{n}{2} + x_{n-1}x_{n-2} > \frac{n}{2} + \frac{(n-1)\sqrt{6}}{6} \cdot \frac{(n-2)\sqrt{6}}{6} = \frac{n^2 + 2}{6} > \frac{n^2}{6},$$
i.e. $x_n > \frac{n\sqrt{6}}{6}$, and this finishes the inductive step.
Next, we prove that $x_n < \frac{n\sqrt{6}}{6} + 1$, for all $n \in \mathbb{N}$. The statement is true for $n=1$ and $n=2$ by inspection. Inequality $(*)$ and the fact $x_{n-1}^2 > \frac{(n-1)^2}{6}$ imply
$$2x_n^2 \le \frac{n^2 + n + 4}{2} - x_{n-1}^2 < \frac{n^2 + n + 4}{2} - \frac{(n-1)^2}{6} = \frac{2n^2 + 5n + 11}{6}.$$
It remains to show
$$\frac{2n^2 + 5n + 11}{6} < 2 \left( \frac{n\sqrt{6}}{6} + 1 \right)^2,$$
but this inequality is equivalent to $5n - 1 < 4n\sqrt{6}$ which holds for $n \ge 3$.