We have given five natural numbers $a,b,c,d,e$ s.t. $(1) \;\; {\textstyle \frac{a}{b} < \frac{c}{d} < \frac{e}{f}}$, $(2) \;\; af - be = -1$. According to (1) we have $ad < bc$ and $cf < de$, i.e. $(3) \;\; ad \leq bc - 1$, $(4) \;\; cf \leq de - 1$. Multiplying (3) and (4) by $f$ and $b$ respectively, we obtain $adf \leq bcf - f$ and $bcf \leq bde - b$, yielding $adf \leq bde - b - f$, i.e. $(be - af)d \geq b + f$, which combined with (2) give us $d \geq b + f$. q.e.d.

be-af=1,bc-ad>0 =>bc-ad>=1 as a,b,c,d are natural numbers similarly de-cf>=1 d=d{be-af}=dbe-daf=dbe-bcf+bcf-daf=b{de-cf}+f{bc-ad}>=b+f d>=b+f