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