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A cyclic problem

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#1 h Sep 5, 2024, 4:54 AM • 1 Y

Y by cubres

Problem. Given non-negative real numbers $a,b,c: ab+bc+ca>0$ then $\frac{1}{a+kb}+\frac{1}{b+kc}+\frac{1}{c+ka}\le f(k)\cdot\frac{a+b+c}{ab+bc+ca}$ where $f(k)=\frac{(k^2-k+1)\left(2k^2+\sqrt{k^2-k+1}+2\sqrt{k^4-k^3+k^2}\right)}{\left(k^2+\sqrt{k^4-k^3+k^2}\right)\left(k^2-k+1+\sqrt{k^4-k^3+k^2}\right)}.$ Also, $k\ge k_{0}\approx 1.874799...$ and $k_{0}$ is largest real root of the equation $k^8 - 3 k^7 + 10 k^6 - 25 k^5 + 30 k^4 - 25 k^3 + 10 k^2 - 3 k + 1=0.$ k=2

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DipoleOfMonorak

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DipoleOfMonorak

#3 h Sep 5, 2024, 4:30 PM

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KhuongTrang wrote:

Problem. Given non-negative real numbers $a,b,c: ab+bc+ca>0$ then $\frac{1}{a+kb}+\frac{1}{b+kc}+\frac{1}{c+ka}\le f(k)\cdot\frac{a+b+c}{ab+bc+ca}$ where $f(k)=\frac{(k^2-k+1)\left(2k^2+\sqrt{k^2-k+1}+2\sqrt{k^4-k^3+k^2}\right)}{\left(k^2+\sqrt{k^4-k^3+k^2}\right)\left(k^2-k+1+\sqrt{k^4-k^3+k^2}\right)}.$ Also, $k\ge k_{0}\approx 1.874799...$ and $k_{0}$ is largest real root of the equation $k^8 - 3 k^7 + 10 k^6 - 25 k^5 + 30 k^4 - 25 k^3 + 10 k^2 - 3 k + 1=0.$ k=2

Let
$\begin{align*} &p_0=1.87479866584... ~\text{is root of polinom}~p^6-2\,p^5+7\,p^4-16\,p^3+7\,p^2-2\,p+1, \\ &F(a,b,c) = \frac{(a\,b+b\,c+c\,a)\Bigl(\bigl(a^2+b^2+c^2\bigr)\,p_0+(a\,b+b\,c+c\,a)\bigl(p_0^2+p_0+1\bigr)\Bigr)}{(a+b+c)(a\,p_0+c)(b\,p_0+a)(c\,p_0+b)} ; \end{align*}$
then min $M$ such that $M\frac{a+b+c}{ab+bc+ca}\geq \frac{1}{a+p_0b}+\frac{1}{b+p_0c}+\frac{1}{c+p_0a}$ equals max $F$

Let
$\begin{align*} &k_0=8.37259964033...~\text{is root of polinom}~k^6-2\,k^5-29\,k^4-156\,k^3-345\,k^2-450\,k-171, \\ &G(a,b,c) = F\bigl(k_0\,b+c, a+b, 0\bigr) - F(a,b,c) \end{align*}$
then $G\left(a,b,u+\frac{a+b}{2}\right)\geq 0$ (proof) $\Rightarrow~F\bigl(k_0\,b+c, a+b, 0\bigr) \geq F(a,b,c)$ for $c=\max(a,b,c)$

$\Rightarrow~F_{max}=\max{\left[F(x,1,0)\right]}=\frac{p_0^3+1-2\,p_0\,\sqrt{p_0^2-p_0+1}}{p_0(p_0-1)^2}$
remark

This post has been edited 1 time. Last edited by DipoleOfMonorak, Sep 5, 2024, 7:17 PM
Reason: remark

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