$a_1=1$; $a_{n+1}=\sqrt{3a_n-1}$ $\quad$ $(n\ge1)$
Now I have to show it is true for $n=1$, which is easy. I have to assume it is true for $n=k$, therefore:
$\sqrt{3a_{k}-1}$ $\gt$ $\sqrt{3a_{k-1}-1}$
And I have to show that it is true for $n=k+1$ , so I have to prove:
$\sqrt{3a_{k+1}-1}$ $\gt$ $\sqrt{3a_{k}-1}$
Have I set this out correctly?
This is the point where I get stuck. From looking at it, it already looks like a very weak induction.
On
First note that $a_n\geq 1\implies a_{n+1}\geq 1. $ So by induction, the positive real number $a_{n+1}=\sqrt {3 a_n-1}$ exists for every $n\in N.$
So the following is valid: $a_{n+2}>a_{n+1}\iff \sqrt {3 a_{n+1}-1}>\sqrt {3 a_n-1}\iff 3 a_{n+1}-1>3 a_n-1\iff a_{n+1}>a_n.$
The second equivalence in the above line is valid because $x\geq 1\implies 3 x-1>0.$
Your inductive hypothesis is: $\sqrt{3a_{k} - 1} > a_{k}$.
In proving the $k \implies k+1$ case:
Remember that $\sqrt{3a_{k+1} - 1}$ can be re-written in terms of $a_k$.
Specifically:
$a_{k+1} = \sqrt{3a_{k} - 1}$; so your inequality becomes:
$$\sqrt{3\sqrt{3a_{k} - 1} - 1} > \sqrt{3a_{k} - 1} \iff 3\sqrt{3a_{k} - 1} - 1 > 3a_{k} - 1 \iff \sqrt{3a_{k} - 1} > a_{k}$$
where we have assumed positivity throughout, and the final inequality is your inductive hypothesis.