TITLE: Proving by induction that the sequence $a_{n+1}=\sqrt{3a_n-1}$ is increasing
QUESTION [5 upvotes]: $a_1=1$;  $a_{n+1}=\sqrt{3a_n-1}$  $\quad$    $(n\ge1)$
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.

REPLY [4 votes]: 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 $x\geq 1\implies 3 x-1>0.$