Prove the recurrence sequence $a_{n+1} = \frac{4 + 3a_{n}}{3 + 2a_{n}} (a_0 = 1)$ to be bounded by $0$ and $\sqrt{2}$ The sequence is defined by: $a_0 = 1$ and $a_{n+1} = \frac{4 + 3a_n}{3 +2a_n}$.
I have to prove that $0 < a_n < \sqrt{2}\,$  holds for any $n \in \mathbb{N}$.
My attempt is to assume $0 < a_n < \sqrt{2}\,$ which is true for $a_0$ at least and a very small $\epsilon$:
Since $\frac{4 + 3(0 + \epsilon)}{3 + 2(0 + \epsilon)} = \frac{4}{3} + \epsilon > 0$ and $\frac{4 + 3(\sqrt{2} - \epsilon)}{3 + 2(\sqrt{2} - \epsilon)} = \sqrt{2} - \epsilon < \sqrt{2}$ and because the sequence is monotonic this holds for larger $\epsilon$ as well. So $0 < a_{n+1} < \sqrt{2}$ and by induction this is true for all $n$.
Now my questions: 1) I have not proven the monotonic bit, and I think I'm supposed to prove it by using the result of the bounding proof. How to get around this? 2) Is this an inductive proof? 
 A: Apparently it involves a tiny trick. I was inspired by squaring the formula for $a_{n + 1}$.  

Use induction on $n$ to prove the claim $ 0 \lt a_n \lt \sqrt 2 $ for each $n \in \Bbb N$. 
The base case $n = 1$ is plain. Suppose the statement is true for an arbitrary $n \in \Bbb N$. Then $0 \lt a_n^2 \lt 2 $. Now verify the following inequalities hold. 
$$ 0 \lt 16 + 24a_n + 9a_n^2 \lt 16 + 24a_n + 18 = 34 + 24a_n  \tag{1}$$
$$ 0 \lt 9 + 12a_n + 4a_n^2 \lt 9 + 12a_n + 8 = 17 + 12a_n \tag{2} $$
Now $\dfrac{(1)}{(2)}$ gives,  
$$ 0 \lt \dfrac{(4 + 3a_n)^2}{(3 + 2a_n)^2} \lt 2 \cdot \dfrac{(17 + 12a_n)}{(17 + 12a_n)} = 2 $$
And hence, $$0 \lt  a_{n+ 1}^2 \lt \sqrt 2$$ 
$\mathscr {Q.E.D.}$
A: Induction: $\;0<a_0<\sqrt2\;$ clearly, so assume for all $\;a_k\;,\;\;1\le k\le n\;$ and prove for $\;a_{n+1}\;$ . Clearly, $\;a_{n+1}>0\;$ (why?), so we shall take care only of the upper bound:
$$a_{n+1}:=\frac{4+3a_n}{3+2a_n}\le\sqrt2\iff4+3a_n<3\sqrt2+2\sqrt2 a_n\iff$$
$$(3-2\sqrt2)a_n<3\sqrt2-4\iff a_n<\frac{3\sqrt2-4}{3-2\sqrt2}=17\sqrt2-24$$
But
$$17\sqrt2-24<\sqrt2\iff 2\sqrt2<3\iff 8<9\;,\;\text{which undoubtedly is true, so we're done}$$
