I can't prove $U_{n} \geq 1$

Question

So we have an iteration: $U_{n+1} = 1 + \frac{1}{U_{n}+1}$ where $n$ is a natural number and $U_{0} = 1$

I have to prove $U_{n} \geq 1$ but the problem is that I just can't.

Here is my attempts:

By using Mathematical induction:

1. For $n = 0$ we find $U_{0} \geq 1$ so we pass to the next step.

2. For $n = n + 1$ we have to find $U_{n+1} \geq 1$ and assume that $U_{n} \geq 1$

So:

$U_{n} + 1 \geq 2$

$\frac{1}{U_{n}+1} \leq \frac{1}{2}$

In the end we find:

$\frac{1}{U_{n}+1}+ 1 \leq \frac{3}{2}$

Which is $U_{n+1} \leq \frac{3}{2}$

As you can see, it's not what I should find at all.

I made other attempts of course like making it all one fraction then making an inequality for both denominators and numerators then multiplying them by each other but it also doesn't work because it still gives me $\leq$ instead of $\geq$

Any help?

EDIT: Okay so here is what I did:

I have to prove $U_{n+1} - 1 \geq 0$

Which means I have to prove $\frac{1}{U_{n}+1} \geq 0$

I guess that I can just say that since $1$ is a natural number and $U_{n} + 1 \geq 0$ then $\frac{1}{U_{n}+1} \geq 0$ which means $U_{n+1} - 1 \geq 0$ because a division of a positive number by a positive number is always a positive number and $U_{n} \geq 1$ is successfully proved...right?

But here is something else that made me super confused, we can't prove it in detail at all because:

$U_{n} + 1 \geq 2$

$\frac{1}{U_{n}+1} \leq 1$ which is still not right because it still doesn't prove $\frac{1}{U_{n}+1} \geq 0$

Can somebody please explain this to me? When I explain it in detail by using inequalities, it doesn't give me the correct answer but when I explain it the other way it's proved.

Answer 1

Starting from $$U_n+1 \ge 2 $$ it looks like after that you got off on the wrong track.

The next deduction is to apply the above inequality, together with the inequality $2 > 0$, and the transitive law of inequality, to deduce that $$U_n+1 > 0 $$ Next you deduce that $$\frac{1}{U_n+1} > 0 $$ The justification for this is the theorem that $x > 0 \implies \frac{1}{x} > 0$.

Finally you obtain $$U_{n+1} = 1 + \frac{1}{U_n+1} > 1 + 0 = 1 $$