For $a,b,c>0.$ Prove$:$ $$ \left\{ \sum\limits_{cyc} \left( ab+{b}^{2}+{c}^{2}+ac \right) \right\}^{4}\geq
27\,{ \sum\limits_{cyc}} \left( ab+{b}^{2}+{c}^{2}+ac \right) ^{3}
\left( c+a \right) \left( a+b \right) $$
I found a SOS proof's for it but very ugly. We have$:$
$$\text{LHS}-\text{RHS}=\sum\limits_{cyc} f(a,b,c) (a-b)^2 \geq 0$$
where $$\begin{align*}
f(a,b,c)&=8\,{a}^{6}+26\,{a}^{5}b+96\,{a}^{4}{b}^{2}+20\,{a}^{4}bc+152\,{a}^{3}{
b}^{3}+130\,{a}^{3}{b}^{2}c\\
&\quad +96\,{a}^{2}{b}^{4}+130\,{a}^{2}{b}^{3}c+
106\,{a}^{2}{b}^{2}{c}^{2}+100\,{a}^{2}{c}^{4}\\
&\quad +26\,a{b}^{5}+20\,a{b}^{
4}c+278\,ab{c}^{4}+8\,{b}^{6}+100\,{b}^{2}{c}^{4} \\
& \geq 0\end{align*}
$$
I hope for an alternative solution without using $uvw.$ Thanks!
Best Answer
A full expanding gives $$\sum_{sym}(4a^8+5a^7b+26a^6b^2-7a^5b^3+22a^4b^4+10a^6bc+45a^5b^2c-16a^4b^3c-36a^4b^2c^2-53a^3b^3c^2)\geq0,$$ which is true by Muirhead.