Given three non-negative numbers $a,b,c$. Prove that $\frac{a+b+c}{k}\geqq\sum\limits_{cyc}\frac{a-b}{b+ k}$ for $k= constant$ so that $k> 0$ . 
Given three non-negative numbers $a, b, c$. Prove that
  $$\frac{a+ b+ c}{k}\geqq \frac{a- b}{b+ k}+ \frac{b- c}{c+ k}+ \frac{c- a}{a+ k}$$
  for $k= constant$ so that $k> 0$ .

For $k= 2$, we can use
$$\frac{a+ b}{2}\geqq \frac{a- b}{b+ 2}$$
but for $k= constant$ ? And I think it would be really really easy if expands. Thanks for help that a lot

Suppose that $a$ is the largest and $c$ the smallest we see that: $$\frac{a} {k} \ge \frac{a-b} {b+k} $$ and $$\frac{b} {k} \ge \frac{b-c} {c+k} $$  and as $c-a$ is not positive we deduce: $$\frac{c} {k} \ge \frac{c-a} {a+k} $$ 
Now add these 3 inequalities .
