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

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

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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
No thanks. But I think it don't deserve that. –  molan May 11 '12 at 16:20

$$0\leq \frac{ab}{a+c}=\frac{b}{1+\frac{c}{a}}\leq b\leq 1$$
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 2:40
It is given that $a>0$ –  Julius May 12 '12 at 5:08