Prove the positivity of a sequence Let $a>0$ a real number and $(u_n)$ the sequence defined by 
$$
u_{n+1} = a - \frac{1}{u_n}\text{ and } u_0 = a.
$$
Question: Determine condition on the value of $a>0$ such that the sequence $(u_n)$ is always positive. 
Attempt: I tried to establish a general formula of $u_n$ in order to set up a condition on $a$, by calculating $u_n$ with some $n$:
$$
u_1 = a - \frac{1}{u_0} = a - \frac 1a = \frac{a^2-1}{a},\\
u_2 = a - \frac{1}{u_1}= a - \frac{a}{a^2-1}=\frac{a^3-2a}{a^2-1},\\
u_3 = a - \frac{1}{u_2} = a - \frac{a^2-1}{a^3-2a}=\frac{a^4-3a^2+1}{a^3-2a},\\
u_4 = a - \frac{1}{u_3} = a - \frac{a^3-2a}{a^4-3a^2+1}=\frac{a^5-4a^3+2a-1}{a^4-3a^2+1}.
$$
But I wasn't successful, it seems that there is no general formula of $u_n$. So I would be appreciate for any suggestion of solution. Thank you!

hint
Let $f (x)=a-1/x $.
$f $ is increasing at $(0,+\infty) $.
$(u_n) $ is monotonic.
$u_1 <u_0$ thus it is strictly decreasing.
the sequences terms are $>0$ if the limit (if it exists) is  $\ge 0$.
the limit $l $ satisfies $l=a-1/l .$
if $a\ge 2$ then
$l=(a\pm\sqrt {a^2-4})/2>0$.
If $a<2$  , $u_2<0$.
So, the condition is $a\ge 2$.
