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 $k> 0$ .

Question

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

Answer 1

By adding $3$ to both sides of the given inequality, it is equivalent to $$\frac{(a+k)+(b+k)+(c+k)}{k}\geq \frac{a+k}{b+k}+\frac{b+k}{c+k}+\frac{c+k}{a+k}.\ \ \ \ \ (1)$$ This is also equivalent to $$\frac{A+B+C}{k}\ge \frac{A}{B}+\frac{B}{C}+\frac{C}{A}\ \ \ \ \ (2)$$ for all $A,B,C\geq k$ and $k>0$ (by setting $A=a+k$, $B=b+k$, and $C=c+k$). But this is again equivalent to $$x+y+z\geq \frac{x}y+\frac{y}z+\frac{z}x\ \ \ \ \ (3)$$ for all $x,y,z\ge 1$ (by setting $x=A/k$, $y=B/k$, and $z=C/k$).

However this is trivial as $$x-x/y=x(1-1/y)\ge 0$$ and similarly $y-y/z\ge 0$ and $z-z/x\ge 0$. Therefore (3) is true (with equality case $x=y=z=1$). Thus (2) is true with equality case $A=B=C=k$. Hence (1) and the required inequality is also true with $a=b=c=0$ being the equality case.

In fact, one can improve this solution: $$\frac{a}{k}-\frac{a-b}{b+k}=(a+k)\left(\frac1k-\frac{1}{b+k}\right)\ge 0,$$ we have two other similar inequalities.

Answer 2

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 .