Proving ${ \left\{\sum \left( ab+{b}^{2}+{c}^{2}+ac \right)\right\} }^{4}\geq 27\,{ \sum} ( ab+{b}^{2}+{c}^{2}+ac ) ^{3} ( c+a) ( a+b) $

Question

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!

Answer 1

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.