Generalization of an Olympiad Inequality

Question

Inspired by this Olympiad inequality I ask the following question for $k > 0$.

Find
$$\max_{a>0,b>,c>0}\Big(\frac{a}{ka + b} + \frac{b}{kb + c} + \frac{c}{kc + a}\Big).$$

Is it $\frac3{k+1}$ when $a=b=c$?

Answer 1

No, it's not $\frac{3}{k+1}$.

Try $k=1$.

In this case we obtain $$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\leq\frac{3}{2},$$ which is wrong for $c\rightarrow0^+$, $a=1$ and $b=\frac{1}{2}$.

For $k\geq2$ the maximum is $\frac{3}{k+1}$.

Indeed, for $a=b=c$ we get a value $\frac{3}{k+1}$ and it's enough to prove that $$\sum_{cyc}\frac{a}{ka+b}\leq\frac{3}{k+1}$$ or $$\sum_{cyc}\left(\frac{a}{ka+b}-\frac{1}{k+1}\right)\leq\frac{3}{k+1}-\frac{3}{k} \tag1$$ or $$\sum_{cyc}\frac{b}{ka+b}\geq\frac{3}{k+1},$$ which follows from C-S: $$\sum_{cyc}\frac{b}{ka+b}=\sum_{cyc}\frac{b^2}{kab+b^2}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(kab+b^2)}$$ $$=\frac{3}{k+1}\frac{(k+1)(a+b+c)^2}{3\sum\limits_{cyc}(kab+a^2)}=\frac{3}{k+1}\frac{(k-2)\sum\limits_{cyc}(a^2-ab)+3\sum\limits_{cyc}(kab+a^2)}{3\sum\limits_{cyc}(kab+a^2)}\geq$$ $$\geq\frac{3}{k+1}\frac{3\sum\limits_{cyc}(kab+a^2)}{3\sum\limits_{cyc}(kab+a^2)}=\frac{3}{k+1}.$$ For $0<k<2$ the supremum is $\frac{2}{k}$, which we can prove by the same way.

Done!