Find the minimal constant $k$ :$\frac{a+k}{a+bc}+\frac{b+k}{b+ca}+\frac{c+k}{c+ab}\ge 3k+2,$ holds $ab+bc+ca=1.$

Question

Problem. Find the minimal constant $k$ sastisfying $$\frac{a+k}{a+bc}+\frac{b+k}{b+ca}+\frac{c+k}{c+ab}\ge 3k+2,$$holds for all $a,b,c: ab+bc+ca=1.$

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I've tried more but there is no good approach. Equality holds at $(0,1,1).$

We can rewrite the OP as $$k\left(\frac{1}{a+bc}+\frac{1}{b+ca}+\frac{1}{c+ab}-3\right)\ge 2-\frac{a}{a+bc}-\frac{b}{b+ca}-\frac{c}{c+ab}.$$ I saw that by Cauchy-Schwarz $$2-\frac{a}{a+bc}-\frac{b}{b+ca}-\frac{c}{c+ab}=\frac{bc}{a+bc}+\frac{ca}{b+ca}+\frac{ab}{c+ab}-1,$$ $$\ge \frac{(ab+bc+ca)^2}{(ab)^2+(bc)^2+(ca)^2+3abc}=\frac{abc(2a+2b+2c-3abc)}{(ab)^2+(bc)^2+(ca)^2+3abc}\ge 0,$$which is true since $a+b+c\ge \sqrt{3}.$

Hence $k=0$ is chosen. For the remain, I think we should consider $$Q=\frac{1}{a+bc}+\frac{1}{b+ca}+\frac{1}{c+ab}-3.$$ If $Q>0$ then the desired $k$ is positive. If $Q<0$ then the desired $k$ is negative.

I stopped here. Is my approach is possible? Hope to see others idea. Thank you.

Answer 1

Some thoughts.

Thanks to above Michael Rozenberg's hint, I write a proof. Hope to see responses.

The original inequality $$\frac{a+k}{a+bc}+\frac{b+k}{b+ca}+\frac{c+k}{c+ab}\ge 3k+2. \tag{*}$$

We can rewrite the $(*)$ as $$k\left(\frac{1}{a+bc}+\frac{1}{b+ca}+\frac{1}{c+ab}-3\right)\ge 2-\frac{a}{a+bc}-\frac{b}{b+ca}-\frac{c}{c+ab}.$$ By using Cauchy-Schwarz $$2-\frac{a}{a+bc}-\frac{b}{b+ca}-\frac{c}{c+ab}=\frac{bc}{a+bc}+\frac{ca}{b+ca}+\frac{ab}{c+ab}-1,$$ $$\ge \frac{(ab+bc+ca)^2}{(ab)^2+(bc)^2+(ca)^2+3abc}=\frac{abc(2a+2b+2c-3abc)}{(ab)^2+(bc)^2+(ca)^2+3abc}\ge 0,$$ which is true since $a+b+c\ge \sqrt{3}.$

We will prove $\boxed{k\ge 0.}$

Hence, it is enough to prove $$\frac{1}{a+bc}+\frac{1}{b+ca}+\frac{1}{c+ab}\ge 3.\tag{1}$$ Proof

Let $a+b+c=3u, ab+bc+ca=3v^2,abc=w^3.$ The $(1)$ turns out $$f(w^3)=-3w^6+A(u,v^2)w^3+B(u,v^2)\ge 0.$$ It is $f''(w^3)=-3<0.$

Hence, $f(w^3)$ is concave which $uvw$ says that we only need to prove $(1)$ is true when $w^3$ get extremal value.

We'll check in two cases

• $w^3=0.$ Let $a=0, bc=1$ and $(1)$ becomes $$\frac{1}{b}+\frac{1}{c}\ge 2,$$which is true by AM-GM.
• Two equal variables. Let $0<b=c=x\le 1; a=\dfrac{1-x^2}{2x}.$ By some calculating works, we can rewrite $(1)$ as $$\frac{(x-1)(x^2 - 2 x - 1)\left(x^4-\dfrac{3x^3}{2}-\dfrac{x^2}{2}+\dfrac{x}{2}-\dfrac{1}{6}\right)}{x}\le 0,$$ which is true by $0<x\le 1.$

Hence, $(1)$ is proven.

In next part of proof, we will find $k\ge 0$ sastifying the following is true $$k\ge \frac{2-\dfrac{a}{a+bc}-\dfrac{b}{b+ca}-\dfrac{c}{c+ab}}{\dfrac{1}{a+bc}+\dfrac{1}{b+ca}+\dfrac{1}{c+ab}-3}.\tag{**}$$ Now, we will prove the claim that: "It is enough to prove $(*)$ when either $abc=0$ or two equal variables.(2)"

Indeed, rewrite $(*)$ as $$g(w^3)=-(3k+2)w^6+A(u,v^2)w^3+B(u,v^2)\ge 0.$$ It is $g''(w^3)=-3k-2<0$ since $k\ge 0.$

Hence, $g(w^3)$ is concave which $uvw$ says that we only need to prove $(*)$ is true when $w^3$ get extremal value.

The claim $(2)$ is proven.

Finally, let $0<b=c=x\le 1; a=\dfrac{1-x^2}{2x}.$ We will find $k_0\ge 0$ such that$$k_0=\text{max} \frac{2-\dfrac{\dfrac{1-x^2}{2x}}{\dfrac{1-x^2}{2x}+x^2}-\dfrac{x}{x+x\dfrac{1-x^2}{2x}}-\dfrac{x}{x+x\dfrac{1-x^2}{2x}} }{\dfrac{1}{\dfrac{1-x^2}{2x}+x^2}+\dfrac{1}{x+x\dfrac{1-x^2}{2x}}+\dfrac{1}{x+x\dfrac{1-x^2}{2x}}-3} , x\in(0;1].$$ (In case $w^3=0$ we get $k\ge 0.$)

By wolfram assiting, we obtain the answer $\boxed{k\ge k_0\approx 1.2364.}$

Notice that

When $k=k_0,$ equality holds also at $(0,0.1671,0.1671).$

About $uvw,$ see also here.

Answer 2

It means, of course, that our variables are non-negative.

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, your inequality is equivalent to $$-(3k+2)w^6+A(u,v^2)w^3+B(u,v^2)\geq0,$$ which is concave function of $w^3$.

Thus, by $uvw$ it's remains to check to cases:

1. $w^3=0$;

2. Two variables are equal.

Can you end it now?