Q:

Inequality for positive functions

I'm saya. Please let me ask a question about an inequality.
Let $f,g$ be a function such that $f,g$ : $\mathbb{R}\to [0,\infty)$ and there exists $C>0$ such  that for all $x \in \mathbb{R}$ $f(x)+g(x)\geq c$
I want to show the following inequality:
\begin{eqnarray*}
\left| \left(\frac{cf}{f+g} \right)(x)-\left(\frac{cf}{f+g} \right)(y) \right| \leq |f(x)-f(y)|+|g(x)-g(y)|\quad(x,y \in \mathbb{R})
\end{eqnarray*} 
My process would be as follows:
\begin{eqnarray*}
\left| \left(\frac{f}{f+g} \right)(x)-\left(\frac{f}{f+g} \right)(y) \right| &=&
 \left| \left(\frac{f}{f+g} \right)(x)-\frac{f(y)}{(f+g)(x)}+\frac{f(y)}{(f+g)(x)} -\left(\frac{f}{f+g} \right)(y) \right|\\
&\leq&\left| \frac{f(x)-f(y)}{(f+g)(x)} \right|+ |f(y)| \left| \frac{1}{(f+g)(x)}- \frac{1}{(f+g)(y)} \right|\\
&\leq& \frac{\left|f(x)-f(y)\right|}{c} + |f(y)| \left| \frac{1}{(f+g)(x)}- \frac{1}{(f+g)(y)} \right|\\
\end{eqnarray*} 
\begin{eqnarray*}
|f(y)| \left| \frac{1}{(f+g)(x)}- \frac{1}{(f+g)(y)} \right|&=&|f(y)| \left| \frac{f(y)-f(x)+g(y)-g(x)}{(f+g)(x)(f+g)(y)} \right|\\
&=&|f(y)| \left| \frac{f(y)-f(x)+g(y)-g(x)}{f(x)f(y)+f(x)g(y)+g(x)f(y)+g(x)g(y)} \right|\\
&\leq&|f(y)| \left| \frac{f(y)-f(x)+g(y)-g(x)}{f(x)f(y)+g(x)f(y)} \right|\\
&=&\left| \frac{f(y)-f(x)+g(y)-g(x)}{f(x)+g(x)} \right|\\
&\leq& \frac{|f(x)-f(y)|}{c}+\frac{|g(x)-g(y)|}{c}
\end{eqnarray*}
I could get the following inequality...
\begin{eqnarray*}
\left| \left(\frac{cf}{f+g} \right)(x)-\left(\frac{cf}{f+g} \right)(y) \right| \leq 2|f(x)-f(y)|+|g(x)-g(y)|\quad(x,y \in \mathbb{R})
\end{eqnarray*} 
What should I modify it? Thanks.

A:

Let $a_1,a_2,b_1,b_2$ be positive and such that $a_1+b_1\geq c>0$, $a_2+b_2\geq c>0$.
We have:
$$\frac{a_1}{a_1+b_1}-\frac{a_2}{a_2+b_2}=\frac{a_1b_2-a_2b_1}{(a_1+b_1)(a_2-b_2)}$$
and 
$$a_1b_2-a_2b_1=(a_1-a_2)b_2+a_2(b_2-b_1)$$
Hence
$$|a_1b_2-a_2b_1|\leq |a_1-a_2|b_2+a_2|b_2-b_1|$$
Now prove that:
$$c |\frac{a_1}{a_1+b_1}-\frac{a_2}{a_2+b_2}|\leq |a_1-a_2|+|b_2-b_1|$$
and then put $a_1=f(x), a_2=f(y), b_1=g(x), b_2=g(y)$.

