$a_1=1$ $\,\,\,\,\,\,\,\,\,\,\,$ $a_{n+1}=3-\frac{1}{a_n}$
then $$a_k\lt3$$ $$-a_k\gt-3$$ $$-\frac{1}{a_k}\lt-\frac13$$ $$3-\frac{1}{a_k}\lt3-\frac13$$ $$a_{k+1}\lt\frac83$$
But i shoud have gotten $a_{k+1}\lt3$ to be able to conclude that $a_k\lt3$
What am i doing here wrong?
Do i need to get exactly $a_k\lt3$ in order to prove that this sequnce is bounded below 3? Because $\frac 83$ is still smaller that 3
You need to prove else that $a_n>0$.
If so then by your work $$a_{k+1}<\frac{8}{3}<3.$$
Also, $a_1<3$, which says $a_n<3$ for all natural $n$ by induction.
The fact that $a_n>0$ we can prove by induction again.
Indeed, we'll prove that even $a_n>\frac{3-\sqrt5}{2}.$
$a_1=1>\frac{3-\sqrt5}{2}$ and if $a_n>\frac{3-\sqrt5}{2}$ we have $$a_{n+1}-\frac{3-\sqrt5}{2}=3-\frac{3-\sqrt5}{2}-\frac{1}{a_n}=$$ $$=\frac{3+\sqrt{5}}{2}-\frac{1}{a_n}=\frac{2}{3-\sqrt5}-\frac{1}{a_n}>0$$ and we are done by induction.