How can we show the inequality $1 \le k \le j-1 $?

Question

We have following boundaries for $k$ and $j$.

$1 \le j \le 2a-2b-1 $ , $1 \le k \le a-b-1 $ and $2k \le j $.
Can we show that $1 \le k \le j-1 $ ?

I understand $k \le \left \lfloor{\frac{j}{2}}\right \rfloor \le a-b-\frac{1}{2} $. I don't know how to obtain $1 \le k \le j-1 $.

Answer 1

So since we know $k \leq \frac{j}{2}$, if we can show that $\frac{j}{2} \leq j-1$, we're done. Notice that this inequality is equivalent to $j \geq 2$. But we already know this, as $k \geq 1$ and $j \geq 2k$.