Having some rational number $0<r<1$, such that $r=\frac{b-2}{b}$ and $b$ is odd, it is expected to find some general lower bound in terms of $k$ for the maximum denominator $n_k$ when we decompose $r$ in egyptian fractions in the following way: $$\frac{b-2}{b}=\frac{1}{n_1}+\frac{1}{n_2}+...+\frac{1}{n_k}$$
Concretely, it would be great to show that $$n_k>\frac{kb}{2}$$
Thanks in advance!
EDIT
As there has been no answer yet, I have checked that if we soften the restriction of the numerator being equal to $b-2$ and we let it be some positive integer $a$, then we can build an egyptian fraction such that $n_k=\frac{b}{p_i}$, being $p_i$ some prime number such that $p_i\mid b$, noticing that $$\frac{p_i+\frac{b}{p_i}}{b}=\frac{1}{\frac{b}{p_i}}+\frac{1}{p_i}$$
For example, setting $b=35$, we could derive that $$\frac{12}{35}=\frac{1}{7}+\frac{1}{5}$$
However, it is impossible to build $\frac{b-2}{b}$ following this method, as $b-2$ is odd and this algorithm yields only even numerators.