<s>[INST] 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:
$\tfrac{16}{6} \leqslant U_{n+1} \leqslant \tfrac{24}{5}$
 [/INST] 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:
$$
U_{n+1} = \frac{8U_n-8}{U_n+2} \ge \frac{8\cdot3-8}{3+2} = \frac{16}{5}
$$
</s>