Show that $\frac{a}{a+1}+\frac{b}{b+1}\geq\frac{a+b}{1+a+b}\geq\frac{c}{1+c}$ whenever $c\leq a+b$ I found a post on ME here.
It sais:

Try to show that $$\frac{a}{a+1}+\frac{b}{b+1}\geq\frac{a+b}{1+a+b}\geq\frac{c}{1+c}$$ whenever $c\leq a+b$

I wasn't able to solve this. How does it work?

We know that,
$$\frac{a}{1+a} \ge \frac{a}{1+a+b} $$   
and 
$$\frac{b}{1+b} \ge \frac{b}{1+a+b} $$ 
Adding the above two equations.
$$\frac{a}{1+a} + \frac{b}{1+b} \ge \frac{a}{1+a+b} + \frac{b}{1+a+b} $$ 
$$\frac{a}{1+a} + \frac{b}{1+b}\ge \frac{a+b}{1+a+b} $$ 
Now assume $a+b = k$
Given
\begin{align}
& a+b \ge c 
\\ &\Rightarrow k \ge c 
\\ &\Rightarrow k + 1 \ge c + 1 
\\ &\Rightarrow \frac{1}{k + 1} \le \frac{1}{c+ 1} 
\\ &\Rightarrow -\frac{1}{k + 1} \ge -\frac{1}{c + 1} 
\\ &\Rightarrow 1-\frac{1}{k + 1} \ge 1 -\frac{1}{c + 1} 
\\ &\Rightarrow \frac{k}{k + 1} \ge \frac{c}{c + 1} 
\\&\Rightarrow \frac{a+b}{1 + a+ b} \ge \frac{c}{c + 1}
\end{align}
Hence Proved
$$\frac{a}{1+a} + \frac{b}{1+b}\ge \frac{a+b}{1+a+b} \ge \frac{c}{c + 1}$$ 
