For what values of $k>0$ does $abc=1 \implies \sum_{\mbox{cyc}}\left(\frac1{a+k}-\frac{a}{a^2+k}\right) \geq 0$?

Question

For what values of $k>0$ does $$abc=1 \implies \left(\frac1{a+k}-\frac{a}{a^2+k}\right) + \left(\frac1{b+k}-\frac{b}{b^2+k}\right)+ \left(\frac1{c+k}-\frac{c}{c^2+k}\right)\geq 0$$

This question arose while I was trying to find a proof of For $abc=1$ prove that $\sum\limits_\text{cyc}\frac{1}{a+3}\geq\sum\limits_\text{cyc}\frac{a}{a^2+3}$

The observation is that for general $k>0$, when $a=b=c=1$, $f(a,b,c;k) \equiv \sum_{\mbox{cyc}}\left(\frac1{a+k}-\frac{a}{a^2+k}\right)=0$, and that the minimum of $f(a,b,c=\frac1{ab};k)$ occurs at a point where at least two of the three variables $a,b,c$ are equal.

However, it is not in general true that $abc=1 \implies f(a,b,c;k)\geq 0$. For example, $$ f(2,2,\frac14;k=5) = -\frac{10}{567} $$ Numerical experimentation tells me that for $\,0<k<4.1649638575$, $abc=1 \implies f(a,b,c;k)\geq 0$, but that for $k>4.1649638575$, there is always some value of $a > 1.69$ such that $$ f(a,a,\frac1{a^2};k) < 0 $$

How would we prove these inequalities (even the $k=3$ case is proving to be quite difficult, yet I "know" that the $k=4$ case is also true)?

And does anybody recognize that upper bound $k>4.1649638575$ such that the implication holds for $k$ below that boundary?

Answer 1

A proof for $k=4$.

Let $a=\frac{y}{x}$, $b=\frac{z}{y}$ and $c=\frac{x}{z}$, where $x$, $y$ and $z$ are positives.

Hence, we need to prove that $$\sum_{cyc}\frac{x}{4x+y}\geq\sum_{cyc}\frac{xy}{4x^2+y^2}$$ or $$\sum_{cyc}\frac{x^3-x^2y}{(4x+y)(4x^2+y^2)}\geq0.$$ Now, let $x=\min\{x,y,z\}$, $y=x+u$ and $z=x+v$.

Hence, we need to prove that $$125(u^2-uv+v^2)x^7+25(10u^3+29u^2v-24uv^2+10v^3)x^6+$$ $$+25(14u^4+55u^3v+27u^2v^2-51uv^3+14v^4)x^5+$$ $$+(225u^5+1279u^4v+1917u^3v^2-842u^2v^3-654uv^4+225v^5)x^4+$$ $$+(25u^6+679u^5v+1560u^4v^2+426u^3v^3-786u^2v^4+71uv^5+25v^6)x^3+$$ $$+uv(76u^5+740u^4v+512u^3v^2-286u^2v^3+59uv^4-v^5)x^2+$$ $$+u^2v^2(84u^4+264u^3v-92u^2v^2+37uv^3+7v^4)x+$$ $$+u^3v^3(32u^3-12u^2v+4uv^2+v^3)\geq0$$ and since $v^6x^3-uv^6x^2+u^2v^6x\geq0$, it remains to prove that: $$125(u^2-uv+v^2)x^7+25(10u^3+29u^2v-24uv^2+10v^3)x^6+$$ $$+25(14u^4+55u^3v+27u^2v^2-51uv^3+14v^4)x^5+$$ $$+(225u^5+1279u^4v+1917u^3v^2-842u^2v^3-654uv^4+225v^5)x^4+$$ $$+(25u^6+679u^5v+1560u^4v^2+426u^3v^3-786u^2v^4+71uv^5+24v^6)x^3+$$ $$+uv(76u^5+740u^4v+512u^3v^2-286u^2v^3+59uv^4)x^2+$$ $$+u^2v^2(84u^4+264u^3v-92u^2v^2+37uv^3+6v^4)x+$$ $$+u^3v^3(32u^3-12u^2v+4uv^2+v^3)\geq0.$$ Let $x=\sqrt{uv}t$.

Hence, $$125(u^2-uv+v^2)x^7+25(10u^3+29u^2v-24uv^2+10v^3)x^6+$$ $$+25(14u^4+55u^3v+27u^2v^2-51uv^3+14v^4)x^5+$$ $$+(225u^5+1279u^4v+1917u^3v^2-842u^2v^3-654uv^4+225v^5)x^4+$$ $$+(25u^6+679u^5v+1560u^4v^2+426u^3v^3-786u^2v^4+71uv^5+24v^6)x^3+$$ $$+uv(76u^5+740u^4v+512u^3v^2-286u^2v^3+59uv^4)x^2+$$ $$+u^2v^2(84u^4+264u^3v-92u^2v^2+37uv^3+6v^4)x+$$ $$+u^3v^3(32u^3-12u^2v+4uv^2+v^3)\geq$$ $$\geq\sqrt{u^9v^9}(125t^7+440t^6+257t^5-136t^4-52t^3+241t^2+155t+10)\geq0,$$ where the last inequality is obviously true.

Done!