Showing a sequence is increasing Given:

How can I show that:
1) If $ \beta >0$ , then the sequence is increasing ? 
2) The sequence converges if and only if $0<\beta \leq 1 $ 
I tried estimating $ a_{n+1} / a_n$ , but without any success. Induction might help , but I can't prove that $a_1 < a_2 $ . 
Will you please help ? 
Thanks in advance 
 A: hint

(1) not by induction and just use the fact $a+b\geq 2\sqrt{ab}$: $a_{n+1}=\frac{1}{2}(\beta a_n^2+\frac{1}{\beta})\geq{a_n}$
(2) you can first consider if the limit exists by assuming $a$: then we will have a equation $a=\frac{1}{2}(\beta a^2+\frac{1}{\beta})$ and this equation must have real number solution

A: Note that
$$\frac{a_{n+1}}{a_n}=\frac{1}{2}\left(\beta a_n +\frac{1}{\beta a_n}\right).$$
Let $x=\beta a_n$. It is a standard fact that for any positive real number $x$, we have $x+\frac{1}{x}\ge 2$. For this inequality is equivalent to $\frac{(x-1)^2}{x}\ge 0$. Equality holds if $x=1$, so if $\beta=1$ our sequence, though non-decreasing, is not strictly increasing. In all other cases, the sequence is strictly increasing.
Now we deal with the convergence question. There is no issue at $\beta=1$. 
First examine the case $0\lt \beta\lt 1$. We show that the sequence is bounded above by $\frac{1}{\beta}$. This can be done by induction. The result holds at $n=1$. Suppose it holds at $n=k$. We show it holds at $n=k+1$.
We have
$$a_{k+1}=\frac{1}{2}\left(\beta a_k^2 +\frac{1}{\beta}\right).$$
By the induction hypothesis, we have $\beta a_k^2\lt \frac{1}{\beta}$, and therefore $a_{k+1}\lt \frac{1}{2}\left(\frac{1}{\beta}+\frac{1}{\beta}\right)=\frac{1}{\beta}$. 
Now suppose that $\beta\gt 1$. If the sequence has a limit, let it be $c$. Then
$$c=\frac{1}{2}\left(\beta c^2 +\frac{1}{\beta}\right).$$
Equivalently,
$$\beta^2 c^2 -2c\beta +1=0,$$
that is, $\beta c=1$. This is impossible, since clearly the limit, if it exists, must be $\ge 1$. 
A: I can clear your doubt in $a_2$>$a_1$.you have $a_2$=0.5($a_1$* $\beta$ + $$\frac{1}{\beta}$$)    which comes out to be           $a_2$=0.5($\beta$+$\frac{1}{\beta}$).Now it is assumed in your first case that $\beta$>0.Now for any positive number,x using AP>GP we get x +1/x is always greater than 2.similarly beta +1/beta is greater than 2.hence u can prove that a2>a1
