How to prove simple arithmetic inequality

Question

While reading over someone else's proof of something, I came across the following inequality that was just implied in a step of their proof.

If $a \in \mathbb{N}_{> 0}$, and $\frac{1}{a + 1} < b \leq \frac{1}{a}$, then $\frac{a+1}{a} \leq 1 + 2b$.

How can this be shown?

I tried the natural way, i.e. $\frac{a+1}{a} = 1 + \frac{1}{a}$ and then $2b > \frac{2}{a+1} \geq \frac{2}{a+a} = \frac{1}{a}$, but this gives me a strict inequality. For the intent of the proof I am going over, I would expect this to be a non-strict inequality... am I doing this the wrong way?

Answer 1

I claim that $\frac{2}{1+a}\ge \frac1a$, to see this we just have to cross multiply and we have

$$2a \ge 1+a$$

which is equivalent to

$$a \ge 1$$

which is a true statement.

Hence $$\frac1a \le \frac2{1+a} < 2b$$ adding $1$ to both sides

$$\frac{a+1}{a}\le 1+2b$$

The inequality is strict but it is alright to write "$\le$".

Answer 2

Note that $$ b> \frac {1}{a+1} \ge \frac{1}{2a} \implies b>\frac {1}{2a} \implies 2b > \frac {1}{a} = \frac{a+1}{a} -1 $$

Thus $$\frac{a+1}{a} \leq 1 + 2b$$