let real numbers $a>0$, and sequence $\{x\_{n}\}$ such
$$x\_{1}=1,ax\_{n}=x\_{1}+x\_{2}+\cdots+x\_{n+1}$$
Find the minimum of the $a$,such $x\_{n}>0,\forall n\ge 1$

My try: since
$$ax\_{n}=x\_{1}+x\_{2}+\cdots+x\_{n+1}$$
$$ax\_{n+1}=x\_{1}+x\_{2}+\cdots+x\_{n+2}$$
$$\Longrightarrow a(x\_{n+1}-x\_{n})=x\_{n+2}$$
then we have
$$r^2-ar+a=0$$
case1: $$\Delta=a^2-4a>0\Longrightarrow a>4$$
then $$r\_{1}=\dfrac{a+\sqrt{a^2-4a}}{2},r\_{2}=\dfrac{a-\sqrt{a^2-4a}}{2}$$
so
$$x\_{n}=Ar^n\_{1}+Br^n\_{2}$$
since
$$x\_{1}=1,x\_{2}=a-1$$
$$\Longrightarrow Ar\_{1}+Br\_{2}=1,Ar^2\_{1}+Br^2\_{2}=a-1$$

I geuss $$a\_{\min}=4$$
then I fell very ugly,Thank you