Prove that $\frac AB \leq \frac{A+C}{B+D} \leq \frac CD$

Question

Assume $\frac{A}{B} \le \frac{C}{D}$ and $A,B,C,D \in \mathbb{R}^+$

Prove that $\frac{A}{B} \le \frac{A+C}{B+D} \le\frac{C}{D}$

(this is a homework exercise I have to do, so please help me :(

Answer 1

In general, when you are asked to prove $m<n$ usually it is easier to start with working out $m-n$. Assuming that $A$,$B$,$C$,$D$ are positive:

$$\frac{A}{B}-\frac{A+C}{B+D}=\frac{AB+AD-AB-BC}{B(B+D)}=\frac{AD-BC}{B(B+D)} = \frac{-BD(\frac{C}{D}-\frac{A}{B})}{B(B+D)}\le0$$

$$\frac{C}{D}-\frac{A+C}{B+D}=\frac{BC+CD-AD-CD}{D(B+D)}=\frac{BC-AD}{D(B+D)}= \frac{BD(\frac{C}{D}-\frac{A}{B})}{B(B+D)}\ge 0$$