# 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 $c \in [0, 1]$?

-

5
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

## 1 Answer

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

-

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
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9267320036888123, "perplexity_flag": "middle"}