Proof that $ab/(a+c)$ is bounded for bounded $b$ and $c$

Question

Is there a concise way to prove that $\frac{ab}{a+c} \in [0, 1]$ for all $a > 0$, $b \in [0, 1]$, and $c \in [0, 1]$?

Note that $0\leq ab\leq a\leq a+c$. Then you can get the conclusion. — molan, May 11 '12 at 16:08
@molan Perfect, thanks! Make it an answer and it gets my vote. — ezod, May 11 '12 at 16:15

Julius · Accepted Answer · 2012-05-11T16:33:27.940

2

$$0\leq \frac{ab}{a+c}=\frac{b}{1+\frac{c}{a}}\leq b\leq 1$$

edited May 11 '12 at 16:33

answered May 11 '12 at 16:26

Julius

You could add $0 \le$ on the left hand side and $\le 1$ on the right – Henry May 11 '12 at 16:31
You also should assume that $a\not=0$. – molan May 12 '12 at 02:40
It is given that $a>0$ – Julius May 12 '12 at 05:08

1 Answers1