Prove that $\left(\frac{a}{b}+ \frac{b}{c}+\frac{c}{a}+x \right) \left(\frac{a^2}{b} + \frac{b^2}{c}+\frac{c^2}{a}+y \right) \ge z$ for $a+b+c=1$

Question

Let $a,b,c>0$ such that $a+b+c=1$.
Prove that for all $m>0$, the following inequality holds: \begin{align*} \left(\frac{a}{b} + \frac{b}{c} + \frac{c}{a} -\frac{m^4+2m^3-8m-4}{m^2} \right) \left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{m^3-m-8}{m} \right)\\ \ge- \frac{\left(m+2 \right) \left(m-2 \right)^2 \left(m+1\right)^2\left(m^2+2m+4\right)}{m^3} \end{align*}

Remark.
For the particular case $m=1$, it is the KaiRain's problem posted at AoPS: \begin{align*} \left(\frac{a}{b} +\frac{b}{c}+\frac{c}{a}+9 \right) \left(\frac{a^2}{b}+\frac{b^2}{c} +\frac{c^2}{a} -8\right) \ge -84 \end{align*}

This is a hard inequality, with the equality at $a=b=c= \frac{1}{3}$ and also for $\left(a,b,c \right)= \frac{4}{7} \left(\sin^2 \frac{3 \pi}{7}, \sin^2 \frac{2 \pi}{7}, \sin^2 \frac{\pi}{7} \right)$ or any cyclic permutations.

A SOS (Sum of Squares) proof was given: \begin{align*} \left(\frac{a}{b} +\frac{b}{c}+\frac{c}{a}+9 \right) \left(\frac{a^2}{b}+\frac{b^2}{c} +\frac{c^2}{a} -8a-8b-8c \right) +84 \left(a+b+c \right) \end{align*} \begin{align*} = \frac{1}{abc\left(a+b+c\right)^2} \left[\sum_{\text{cyc}} \frac{\left(a^3c+3a^2bc-a^2c^2-2ab^3-2abc^2+b^3c-b^2c^2+bc^3 \right)^2}{bc } \\+ \sum_{\text{cyc}} \frac{ \left(a^2b+3a^2c-3ab^2-2ac^2+b^2c \right)^2}{2 } \right]\ge 0 \end{align*}

Also, the generalization can be expressed in SOS form, but it is a bit difficult. Some base method likes $uvw$ or $pqr$,... may lead to high degree, I try to use but without success. For each value of $m \neq 2$ there are four cases of equality, which is the hardest part of the solution.

Is there any good way to deal with it?

Answer 1

Here is an SOS (Sum of Squares) solution.

We have \begin{align*} &\Big(\frac{a}{b} + \frac{b}{c} + \frac{c}{a} -\frac{m^4+2m^3-8m-4}{m^2} \Big) \Big(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{m^3-m-8}{m} (a + b + c) \Big)\\[6pt] &\qquad + \frac{\left(m+2 \right) \left(m-2 \right)^2 \left(m+1\right)^2\left(m^2+2m+4\right)}{m^3}(a+b+c)\\[6pt] ={}& \frac{1}{m^2(a+b+c)a^2b^2c^2} \left[\sum_{\mathrm{cyc}} \frac34ac(a^2bm - a^2cm + ab^2m - b^2cm - 2ab^2 + 2bc^2)^2\right.\\[6pt] &\qquad \left. + \sum_{\mathrm{cyc}} g(a,b,c, m) + h(a,b,c,m)\right.\\ &\qquad \left. + \frac34 a^2(-abcm^2 + b^2cm^2 + a^2cm - b^3m + 2b^2c - 2bc^2)^2\right] \end{align*} where \begin{align*} g(a, b, c, m) &:= \frac14 ac\left(-2ab^2m^2 + 2b^2cm^2 + a^2bm + a^2cm + ab^2m\right.\\ &\qquad\qquad \left. - 2abcm + b^2cm - 2bc^2m - 2ab^2 + 4abc - 2bc^2\right)^2, \\ h(a,b,c,m) &:= \frac14 \left(-a^2bcm^2 - ab^2cm^2 + 2abc^2m^2 + a^3cm + ab^3m \right.\\ &\qquad \left. - 2bc^3m - 4a^2bc + 2ab^2c + 2abc^2\right)^2. \end{align*}

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Remark. Here is the Maple expression inside the square bracket (if someone wants to verify the above identity):

(3*a^2*(- a^2*c*m + a*b*c*m^2 + b^3*m - b^2*c*m^2 - 2*b^2*c + 2*b*c^2)^2)/4 + (a^3*c*m - a^2*b*c*m^2 - 4*a^2*b*c + a*b^3*m - a*b^2*c*m^2 + 2*a*b^2*c + 2*a*b*c^2*m^2 + 2*a*b*c^2 - 2*b*c^3*m)^2/4 + (3*a*c*(2*a*b^2 - 2*b*c^2 - a*b^2*m - a^2*b*m + a^2*c*m + b^2*c*m)^2)/4 + (3*b*c*(2*a*b^2 - 2*a^2*c - a^2*b*m + a*c^2*m + a^2*c*m - b*c^2*m)^2)/4 + (3*a*b*(2*a^2*c - 2*b*c^2 - a*b^2*m - a*c^2*m + b*c^2*m + b^2*c*m)^2)/4 + (b*c*(2*a^2*b*m^2 + a^2*b*m - 2*a^2*c*m^2 + a^2*c*m - 2*a^2*c - 2*a*b^2*m - 2*a*b^2 - 2*a*b*c*m + 4*a*b*c + a*c^2*m + b*c^2*m)^2)/4 + (a*c*(a^2*b*m + a^2*c*m - 2*a*b^2*m^2 + a*b^2*m - 2*a*b^2 - 2*a*b*c*m + 4*a*b*c + 2*b^2*c*m^2 + b^2*c*m - 2*b*c^2*m - 2*b*c^2)^2)/4 + (a*b*(- 2*a^2*c*m - 2*a^2*c + a*b^2*m - 2*a*b*c*m + 4*a*b*c + 2*a*c^2*m^2 + a*c^2*m + b^2*c*m - 2*b*c^2*m^2 + b*c^2*m - 2*b*c^2)^2)/4