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

$$\begin{array}{rl}2x_n^2& \le x_{n-1}^2+x_{n-2}^2+n,\\ 2x_{n-1}^2& \le x_{n-2}^2+x_{n-3}^2+n-1,\\ & ⋮\\ 2x_3^2& \le x_2^2+x_1^2+3,\\ 2x_2^2& \le x_1^2+x_0^2+2.\end{array}$$ Adding all these inequalities gives $$2x_n^2+x_{n-1}^2\le 2x_1^2+x_0^2+[n+(n-1)+\cdots +3+2],$$ i.e. $$2x_n^2+x_{n-1}^2\le \frac{n^2+n+4}{2}.(\ast )$$ If there exists a sought number $A$, the left hand side of this inequality is at least $2A^2{n}^2+A^2(n-1{)}^2$, thus $$2A^2{n}^2+A^2(n-1{)}^2\le \frac{n^2+n+4}{2},$$ i.e. $6A^2{n}^2-4A^{2}n+2A^2\le n^2+n+4$. 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 $(\ast )$ 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 $n\ge 3$.