Why are these two inequalities not the same even though they use the same equation? I just don't get it, like at all.
$U_{n}$ is an iteration defined on $\mathbb{N}$, BTW.
The question was:

$$\begin{align}
U_{n+1} &=
\frac{8U_n - 8}{U_{n} + 2} = 8 - \frac{24}{U_{n} + 2}\\
U_0 &= 3
\end{align}
$$
"Prove that $3 \leqslant U_n \leqslant 4$ by using mathematical induction"

Step 1: Test $n = 0$, yes, it's correct.
Step 2: Let's say the inequality is correct and test it for $n+1$
I used the first equation BTW. The result(for me) was:
$\tfrac{16}{6} \leqslant U_{n+1} \leqslant \tfrac{24}{5}$
Now, the exercise solved it by using the second equation, and its result was:
$\tfrac{16}{5} \leqslant U_{n+1} \leqslant 4 $
And since $16/5$ is bigger than 3 then it's correct and works.
How are there two different results and inequalities when you use each one of them?
Isn't $\tfrac{8U_{n} - 8}{U_{n} + 2}$ supposed to be equal to $8 - \tfrac{24}{U_{n} + 2}$ which means no matter which one you use, and there will be the same result?
I just don't get it. Which one is correct?
How are there two different results and inequalities when you use each one of them?
My solution:
An inequality for the numerator:
\begin{align}
3 &\leqslant U_n \leqslant 4\\
24 &\leqslant U_n \leqslant 32\\
16 &\leqslant U_n \leqslant 24
\end{align}
Now the equality for the dominator:
\begin{align}
3 &\leqslant U_n \leqslant 4\\
5 &\leqslant U_n + 2 \leqslant 6\\
\tfrac{1}{6} &\leqslant \tfrac{1}{U_{n} + 2} \leqslant \tfrac{1}{5}
\end{align}
By multiplying each side of the inequality then:
$\tfrac{16}{6} \leqslant U_{n+1} \leqslant \tfrac{24}{5}$
 A: I think I see what is going on. Let's focus on your left inequality. You basically do this:
$$
U_{n+1} = \frac{8U_n-8}{U_n+2} \ge \frac{8\cdot3-8}{U_n+2}\ge \frac{8\cdot3-8}{4+2} = \frac{16}{6}
$$
This is absolutely correct. However, we lose some tightness in the bound, because $\frac{8U_n-8}{U_n+2}$ is in fact increasing in $U_n$ (for $U_n>-2$). This is not obvious, but can be seen immediately in the rewritten form $8-\frac{24}{U_n+2}$. Therefore, you can replace $U_n$ with $3$ everywhere to get:
$$
U_{n+1} = \frac{8U_n-8}{U_n+2} \ge \frac{8\cdot3-8}{3+2} = \frac{16}{5}
$$
A: The book used the second inequality, and gets a tigher result (stronger bounds) than you. Why?
This is because, using the first expression, you bound separately numerator and denominator, and that's not the optimal thing to do. Namely, you known $m \leq u_n \leq M$, and then bound
$$
\frac{8 m-8}{M+2} \leq \frac{8 u_{n}-8}{u_n+2}  \leq \frac{8 M-8}{m+2}  \tag{1}
$$
This is correct, but not optimal. You can't see immediately it from the expression you use, but using the second expression is equivalent to bounding it as
$$
\frac{8 m-8}{m+2} \leq \frac{8 u_{n}-8}{u_n+2}  \leq \frac{8 M-8}{M+2}  \tag{2}
$$
which is better (but, again, it is not not obvious you can do this based on the first expression$^{\dagger}$).

${}^{(\dagger)}$ a way to realize you can indeed do this without using the second expression (as the book did) is to notice that the function $f\colon x\to \frac{8x-8}{x+2}$ is increasing, and so $f(m)\leq f(u_n) \leq f(M)$ whenever $m\leq u_n\leq M$.
A: You are not being careful with inequalities. Note that
$$4=\frac{24}6 \le \frac{24}{U_n+2}\le \frac{24}5,$$
so 
$$8-\frac{24}5 \le 8-\frac{24}{U_n+2} \le 8-4 .$$
(Both taking the reciprocal and taking the additive inverse reverse inequalities.)
