Erin Lozano

2022-06-25

Prove that $\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$
I'm reading a introductory book on mathematical proofs and I am stuck on a question.
Let $a,b,c,d$ be positive real numbers, prove that if $\frac{a}{b}<\frac{c}{d}$, then $\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$

kpgt1z

Expert

HINT: Note that
$\begin{array}{rl}\frac{a+c}{b+d}-\frac{a}{b}& =\frac{a+c}{b+d}\cdot \frac{b}{b}-\frac{a}{b}\cdot \frac{b+d}{b+d}\\ & =\frac{b\left(a+c\right)-a\left(b+d\right)}{b\left(b+d\right)}\\ & =\frac{ab+bc-ab-ad}{b\left(b+d\right)}\\ & =\frac{bc-ad}{b\left(b+d\right)}\phantom{\rule{thickmathspace}{0ex}}.\end{array}$
You’d like to show that this difference is positive, so you’d like to know that $bc>ad$. Use the fact that $\frac{a}{b}<\frac{c}{d}$ to show this.
The other inequality can be handled similarly.

arridsd9

Expert

$\frac{a}{b}=\frac{a\left(1+\frac{d}{b}\right)}{b\left(1+\frac{d}{b}\right)}=\frac{a+d\left(\frac{a}{b}\right)}{b+d}<\frac{a+d\left(\frac{c}{d}\right)}{b+d}=\frac{a+c}{b+d}$
and similarly for the other inequality.

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