Is there a general relation between $a/b$ and $(a+c)/(b+c)$ where $a,b,c > 0 $? Is there a general relation between $a/b$ and $(a+c)/(b+c)$ where $a,b> 0$ and $c\geq 0$ ? 
Is there a general proof for that relation ? 
 A: If $\underline{a\ge b}$ then $ac\ge bc$ hence $ab+ac=a(b+c)\ge ab+bc=b(a+c)$ so 
$$\frac ab\ge \frac{a+c}{b+c}$$
A: Good observation, these inequalities are quite useful. But you need a little bit more:


*

*if $a\ge b$, then $\dfrac ab\ge\dfrac{a+c}{b+c}$

*if $a\le b$, then $\dfrac ab\le\dfrac{a+c}{b+c}$


You can prove it by multiplying by the common denominator:


*

*$a(b+c)\ge b(a+c)\Longleftrightarrow ac\ge bc$

*$a(b+c)\le b(a+c)\Longleftrightarrow ac\le bc$

A: $\frac{a+c}{b+c}$ is between $\frac ab$ and $\frac cc=1$, hence whether $\frac{a+c}{b+c}>\frac ab$ or $\frac{a+c}{b+c}<\frac ab$ (or $\frac{a+c}{b+c}=\frac ab$) depends on how $\frac ab$ compares to $1$, i.e. how $a$ compares to $b$.
A: Suppose $a,b,c \in \mathbb N$, $a < b$.
$$\color{red}{\dfrac ab} 
  = \dfrac{a(b+c)}{b(b+c)} 
  = \dfrac{ab+ac}{b(b+c)}
  \color{red}{<} \dfrac{ab+bc}{b(b+c)} 
  = \dfrac{b(a+c)}{b(b+c)} 
  = \color{red}{\dfrac{a+c}{b+c}}
  \color{red}{<} \dfrac{b+c}{b+c} = \color{red}1$$
Suppose $a,b,c \in \mathbb N$, $a > b$.
$$\color{red}{\dfrac ab} 
  = \dfrac{a(b+c)}{b(b+c)} 
  = \dfrac{ab+ac}{b(b+c)}
  \color{red}{>} \dfrac{ab+bc}{b(b+c)} 
  = \dfrac{b(a+c)}{b(b+c)} 
  = \color{red}{\dfrac{a+c}{b+c}}
  \color{red}{>} \dfrac{b+c}{b+c} = \color{red}1$$
