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 
 A: you've done a good work, do not feel bad about it and think your guess is also true( i havent' checked it yet).
If you continue , you will reach the answer eventually.

However, this way of approach forces us to do many caculations. 
I would like to present to you the below problem and its proof to reduce your work:
Problem
 $\{x_n\}$ is positive real sequence and $a$ is a positive number .$a<4$.
Then there exists $N \in \mathbb{N}$ such that:
$ x_1+x_2+..+x_N+x_{N+1} > a. x_N$.
Proof
Assume that $ \forall n \in \mathbb{N}: x_1+x_2+..+x_n+x_{n+1} \le a. x_n$(1)
Denote $X_n=\sum_{k=1}^n x_k \Rightarrow$ $\{X_n\}$ is increasing sequence
$(1) \Leftrightarrow X_{n+1} \le a( X_n-X_{n-1}) \forall n \ge 2$(2)
Denote $a_n=\frac{X_{n+1}}{X_n} \forall n\ge 1 \Rightarrow a_n \ge 1 \forall n$
$(2) \Leftrightarrow a_n \le a( 1-\frac{1}{a_{n-1}}) \forall n \ge 2$
Thus,$ a \ge a_n+\frac{a}{a_{n-1}} \ge 2\sqrt{\frac{a.a_n}{a_{n-1}}} > a.\sqrt{\frac{a_n}{a_{n-1}}} \forall n \ge 2$ (due to $a<4$)
Therefore$a_{n-1} > a_n \forall n \ge 2$.

In sum, we have $a_n$ is a decreasing sequence and $a_n >1 \forall n$, so it has limit point.
Denote :$ A=\lim_{n \rightarrow \infty} a_n$.
We have:$a_n \le a( 1-\frac{1}{a_{n-1}}) \forall n \ge 2 \Rightarrow A \le a( 1-\frac{1}{A})$.
$\Leftrightarrow a \ge A+\frac{a}{A} \ge 2\sqrt{a}$
 Which is wrong because $a<4$.


Back to your problem, if you use this problem, you only have to prove with the case $a\ge 4$. Hope to keep doing good work
A: 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.
