Prove $13\sum(a+b)^5 \geq 16 \sum ab(a+b)(4a^2+4b^2+4ab+c^2)$

Question

Given $a,b,c>0$, prove that $$13[(a+b)^5+(b+c)^5+(c+a)^5] \geq 16[ab(a+b)(4a^2+4b^2+4ab+c^2)+bc(b+c)(4b^2+4c^2+4bc+a^2)+ca(c+a)(4c^2+4a^2+4ca+b^2)]$$

I tried subtracting the RHS from the LFS but the resulting algebraic expression was too long to handle. I also attempted to see if $16[ab(a+b)(4a^2+4b^2+4ab+c^2)\leq some f(a+b)$ but the $c^2$ weren't helping since there's no relation between $a,b,c.$

Answer 1

$$13\sum_{cyc}(a+b)^5-16\sum_{cyc}ab(a+b)(4a^2+4b^2+4ab+c^2)=$$ $$=13\sum_{_cyc}(2a^5+5a^4b+5a^4c+10a^3b^2+10a^3c^2)-$$ $$-16\sum_{cyc}(4a^4b+4a^4c+4a^3b^2+4a^3c^2+4a^3b^2+4a^3c^2+2a^2b^2c)=$$ $$=\sum_{cyc}(26a^5+a^4b+a^4c+2a^3b^2+2a^3c^2-32a^2b^2c)\geq0,$$ which is true by Muirhead.

Answer 2

Buffalo Way helps! Not nice, but it's a solution. (Sorry for my bad solution)

Let $a=c+x,b=c+y$. We need to prove:

$\left( 224\,{x}^{2}-224\,yx+224\,{y}^{2} \right) {c}^{3}+8\, \left( x +y \right) \left( 34\,{x}^{2}-43\,yx+34\,{y}^{2} \right) {c}^{2}+ \left( 132\,{x}^{4}+8\,{x}^{3}y-20\,{x}^{2}{y}^{2}+8\,x{y}^{3}+132\,{ y}^{4} \right) c+ \left( x+y \right) \left( 26\,{x}^{4}-25\,{x}^{3}y+ 27\,{x}^{2}{y}^{2}-25\,x{y}^{3}+26\,{y}^{4} \right)\geqq 0$

Which is obvious because: $$132\,{x}^{4}+8\,{x}^{3}y-20\,{x}^{2}{y}^{2}+8\,x{y}^{3}+132\,{y}^{4}$$ $$=4 \,yx \left( 134\,{x}^{2}-203\,yx+134\,{y}^{2} \right) +132\, \left( x- y \right) ^{4} \geqq 0$$ And: $$26\,{x}^{4}-25\,{x}^{3}y+27\,{x}^{2}{y}^{2}-25\,x{y}^{3}+26\,{y}^{4}=y x \left( 79\,{x}^{2}-129\,yx+79\,{y}^{2} \right) +26\, \left( x-y \right) ^{4} \geqq 0$$ Done.