Prove that $\frac{a}{b} < \frac{a+c}{b+d} <\frac{ c}{d}$

Question

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 $\frac{a}{b} < \frac{a+c}{b+d} <\frac{ c}{d}$.

Answer 1

HINT: Note that

$$\begin{align*} \frac{a+c}{b+d}-\frac{a}b&=\frac{a+c}{b+d}\cdot\frac{b}b-\frac{a}b\cdot\frac{b+d}{b+d}\\ &=\frac{b(a+c)-a(b+d)}{b(b+d)}\\ &=\frac{ab+bc-ab-ad}{b(b+d)}\\ &=\frac{bc-ad}{b(b+d)}\;. \end{align*}$$

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.

Answer 2

Another hint: For positive numbers, $\; \dfrac ab<\dfrac cd\iff ad<bc$.

Answer 3

$$\frac{a}{b}=\frac{a(1+\frac{d}{b})}{b(1+\frac{d}{b})}=\frac{a+d(\frac{a}{b})}{b+d}<\frac{a+d(\frac{c}{d})}{b+d}=\frac{a+c}{b+d}$$

and similarly for the other inequality.