# Bounds on Egyptian fractions

Having some rational number $$0, such that $$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 $$n_k>\frac{kb}{2}$$

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.