How find this minimum of the $a$ such $x_{n}>0$ 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 

First, we notice that $x_2=a-1$, and hence $a> 1$.
As you noticed, a necessary condition is that the sequence $(x_n)$ is increasing (since $a(x_{n+1}-x_n)$ has to be positive and $a> 1$). Consequently $x_{n+2}=a(x_{n+1}-x_n)> x_{n+1}$ implies $x_{n+1}> \frac{a}{a-1}x_n>\frac{a^n}{(a-1)^n}$. In particular $x_2=a-1>a/(a-1)$, i.e. $(a-1)^2>a$. Also $x_3=a(a-2)>a^2/(a-1)^2$ yields a quartic equation which has a unique root $a_0\approx 2.839$ which moreover satisfies $(a_0-1)^2>a_0$.
Therefore, a necessary condition is that $a>a_0$. 
An induction proves that this is sufficient.
