Proof by induction of a recursive sequence I am studying CIE A levels Further Maths and I am stuck at a question from June 2002:

Q
The sequence of positive numbers $u_1,u_2,u_3,...$ is such that $u_1<4$ and $$u_{n+1}=\frac{5u_n+4}{u_n+2}$$By considering $4-u_{n+1}$, or otherwise, prove by induction that $u_n<4$ for all $n\geq1$.
Prove also that $u_{n+1}>u_n$ for all $n\geq 1$.

Here is my solution
To Prove the Initial Relation $u_n<4$
By using long division:
$$\frac{5u_n+4}{u_n+2}\equiv 5-\frac{6}{u_n+2}$$
For $n=1$
$$u_{2}= 5-\frac{6}{u_1+2}$$
Since $u_1<4$ then,
$$\frac{6}{u_1+2}>1$$
and,
$$5-\frac{6}{u_1+2}<4$$
therefore,
$$u_{2}<4$$
thus the statement is true for $n=1$
Conjecture: The statement is true for $n=k$
$$u_{k}<4$$
Since $u_{k+1}<4$ ... (proved above for $n=1$),
then,
$$\frac{6}{u_{k+1}+2}>1$$
and,
$$5-\frac{6}{u_{k+1}+2}<4$$
therefore,
$$u_{k+2}<4$$
Since the statement is true for $n=k$, $n=k+1$, and $n=1$, the given relation ($u_n<4$) is correct.

To prove that $u_{n+1}>u_n$:
$$u_{n+1}-u_n=\frac{5u_n+4}{u_n+2}-u_n\equiv -\frac{(u_n)^2+3u_n+4}{u_2+2}$$
$$-\frac{(u_n)^2+3u_n+4}{u_2+2}\equiv \frac{(4-u_n)(u_n+1)}{u_n+2}>0$$
Since,
$$u_{n+1}-u_n>0$$
$$u_{n+1}>u_n$$


I am not sure if my solution to the first part is correct, and even if it is correct, I didn't use the relation $4-u_{n+1}$ so it would be really great if someone suggests abetter solution using the relation mentioned in the question.
As for second part, I am pretty confident that it is correct, but I am open to suggestions.

For the record, this is a Ricatti recurrence, as $5 \cdot 2 - 1 \cdot 4 \ne 0$. There are at least three solution ideas around, I like Mitchell's "An Analytic Riccati solution for Two-Target Discrete-Time Control", Journal of Economic Dynamics and Control 24(4), pp 615-622 (2000).
Take:
$$
u_{n + 1} = \frac{a u_n + b}{c u_n + d}
$$
Define a new variable by:
$$
x_n = \frac{1}{1 + \eta u_n}
$$
Substituting into the recurrence gives:
$$
x_{n + 1}
      = \frac{(d \eta - c) x_n + c}
             {(b \eta^2 - (a - d) \eta - c) x_n + a \eta + c}
$$
Selecting $\eta$ such that $b \eta^2 - (a - d) \eta - c = 0$ you get a linear first-order recurrence. Here:
$$
4 \eta^2 -3 \eta - 1 = 0
$$
gets $\eta = 1$ or $\eta = -1/4$. First is simpler, pick that one:
$$
x_{n + 1} = \frac{(2 \cdot 1 - 1) x_n + 2}{5 \cdot 1 + 1} = \frac{x_n + 1}{6}
$$
We just have a limit on $u_1$, but we can just shift by one without changing anything. We get:
$$
x_1 = \frac{1}{1 + u_1}
$$
Use generating functions to solve the linear recurrence by defining $X(z) = \sum_{n \ge 0} x_{n + 1} z^n$, multiplying by $z^n$ and summing over $n \ge 0$:
$$
\frac{X(z) - x_1}{z} = \frac{X(z)}{6} + \frac{1}{6} \cdot \frac{1}{1 - z}
$$
to get:
$$
X(z) = \frac{1}{5 (1 - z)} + \frac{5 x_1 - 1}{5} \cdot \frac{1}{1 - z/6}
$$
The coefficients are immediate:
$$
x_{n + 1} = \frac{1}{5} + \frac{5 x_1 - 1}{5} \cdot 6^{-n}
$$
Going back to the original variables:
\begin{align}
u_n 
  &= \frac{4 (u_1 + 1) \cdot 6^n + (u_1 - 4)}
          {(u_1 + 1) \cdot 6^n - (u_1 - 4)}
  &= 4 - \frac{5 (4 - u_1)}{(u_1 + 1) \cdot 6^n + (4 - u_1)}
\end{align}
If $u_1 < 4$, the second term is always positive, and $u_n < 4$. In that case, the second term decreases as $n$ increases, so $u_{n + 1} > u_n$.
