A nontrvial inequality $\sum\limits_{\rm cyc}a^2 c(4a-3b-c)^2\ge 20abc\sum\limits_{\rm cyc}a(a-b)$

Question

For real numbers $a,b,c>0$, show that $$\sum_{\rm cyc}a^2 c(4a-3b-c)^2\ge 20abc(a^2+b^2+c^2-ab-bc-ca)$$

This has the following dumbass notation: $$\begin{array}{ccccccccccc} &&&&&\cdot&&&&&\\ &&&&\cdot&&16&&&&\\ &&&1&&-44&&-8&&&\\ &&-8&&35&&35&&1&&\\ &16&&-44&&35&&-44&&\cdot&\\ \cdot&&\cdot&&1&&-8&&16&&\cdot \end{array}\ge 0$$

Exact equality takes place at cyclic permutations of $(1:1:1)$, $(4:1:0)$, and $(X:Y:1)$ where $X,Y$ are the largest real roots of $32X^3-132X^2+123X-32$ and $32Y^2-123Y^2+132Y-32$, and the domain cannot be extended to arbitrary real numbers.

So I saw this from social media and I cannot find a proof for this, as SOS did not take me very far. Any help will be appreciated!

Answer 1

We need to prove that: $$\sum_{cyc}(16a^4c+a^3b^2-8a^3c^2-44a^3bc+35a^2b^2c)\geq0$$ or $$\sum_{cyc}(32a^4c+2a^3b^2-16a^3c^2-88a^3bc+70a^2b^2c)\geq0$$ or $$\sum_{cyc}(16a^4b+16a^4c-7a^3b^2-7a^3c^2-88a^3bc+70a^2b^2c)\geq$$ $$\geq\sum_{cyc}(16a^4b-16a^4c-9a^3b^2+9a^3c^2)$$ or

$$\sum_{cyc}(16a^4b+16a^4c-7a^3b^2-7a^3c^2-88a^3bc+70a^2b^2c)\geq$$ $$\geq(a-b)(a-c)(b-c)\sum_{cyc}(16a^2+7ab).$$ We'll prove that:

$$\sum_{cyc}(16a^4b+16a^4c-7a^3b^2-7a^3c^2-88a^3bc+70a^2b^2c)\geq0.$$ Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$,$abc=w^3$ and $u^2=tv^2$.

Thus, $t\geq1$ and we need to prove that $$1296u^3v^2-1485uv^4-810u^2w^3+999v^2w^3\geq0,$$ which is a linear inequality of $w^3$,

which says that it's enough to prove the last inequality for an extreme value of $w^3$,

which happens in the following cases.

1. $w^3\rightarrow0^+$.

Let $c\rightarrow0^+$ and $b=1$.

Thus, we need to prove that $$16a^4+16a-7a^3-7a^2\geq0$$ or $$16a^2-23a+16\geq0,$$ which is obvious;

2. Two variables are equal.

Let $b=c=1$.

Thus, we need to prove that: $$32a^4+32a+32-14a^3-14a^2-14-88a^3-176a+140a^2+70a\geq0$$ or $$(a-1)^2(16a^2-19a+9)\geq0,$$ which is true because $$19^2-4\cdot16\cdot9<0.$$ Thus, it's enough to prove that $$1296u^3v^2-1485uv^4-810u^2w^3+999v^2w^3\geq(a-b)(a-c)(b-c)(144u^2-75v^2)$$ or $$9(48u^3v^2-55uv^4-30u^2w^3+37v^2w^3)\geq(a-b)(a-c)(b-c)(48u^2-25v^2),$$ for which it's enough to prove that $$81(48u^3v^2-55uv^4-30u^2w^3+37v^2w^3)^2\geq(a-b)^2(a-c)^2(b-c)^2(48u^2-25v^2)^2$$ or $$3(48u^3v^2-55uv^4-30u^2w^3+37v^2w^3)^2\geq(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)(48u^2-25v^2)^2$$ or $$(1251u^4-2265u^2v^2+1183v^4)w^6+$$ $$+2u(1152u^6-4008u^4v^2+4682u^2v^4-1995v^6)w^3+v^6(12u^2-25v^2)^2\geq0.$$ Now, since $$1251u^4-2265u^2v^2+1183v^4>0$$ and $$(12u^2-25v^2)^2\geq0,$$ it's enough to prove our inequality for $$1152t^3-4008t^2+4682t-1995<0,$$ which gives $$1\leq t\leq1.6574...$$ and it remains to prove that: $$t(1152t^3-4008t^2+4682t-1995)^2-(1251t^2-2265t+1183)(12t-25)^2\leq0$$ or $$(7-4t)(t-1)^2(12t-13)^2(48t-25)^2\geq0,$$ which is true for $$1\leq t\leq1.6574...$$

Answer 2

This answer is not as apt as Michael Rozenberg's as it is found by computer software (not mine).

The following identity holds:

$$[4(a^2+b^2+c^2)+17(ab+bc+ca)](LHS-RHS) =$$ $$ 14abc\sum_{cyc}(4a^2-4b^2-3ab+8bc-5ca)^2+$$ $$\sum_{cyc}a(2 a^2 b + 12 a^2 c - 6 a b^2 - 10 a b c - 3 a c^2 - 8 b^3 + 25 b^2 c - 12 b c^2)^2$$

Answer 3

Let $f(a, b, c) = \mathrm{LHS} - \mathrm{RHS}$. Suppose the inequality is true. Then $f(x_1^2, x_2^2, x_3^2)\ge 0$ for all real numbers $x_1, x_2, x_3$. We wish $f(x_1^2, x_2^2, x_3^2) = \sum_{i=1}^n f_i^2$ where $f_i$'s are all polynomials. This may be impossible. However, from Artin's theorem, there exists nonzero $g\in \mathbb{R}[x_1, x_2, x_3]$ such that $g^2f = \sum_{i=1}^n f_i^2$. A possible choice is $g = x_1^2 + x_2^2 + x_3^2$. It works. $(x_1^2 + x_2^2 + x_3^2)^2f(x_1^2, x_2^2, x_3^2)$ is SOS (Sum of Squares).

We obtain the following SOS expression. \begin{align} &64(a+b+c)^2(\mathrm{LHS} - \mathrm{RHS})\\ = \ &\sum_{\mathrm{cyc}} 15 a (8 a^2 c+2 a b^2-13 a b c-2 a c^2-8 b^3+17 b^2 c-4 b c^2)^2\\ &\quad + \sum_{\mathrm{cyc}} a (8 a^2 b+24 a^2 c-30 a b^2-a b c-6 a c^2-8 b^3+49 b^2 c-36 b c^2)^2\\ &\quad +174 a b c (4 a^2-3 a b-5 a c-4 b^2+8 b c)^2\\ &\quad +58 a b c (4 a^2-13 a b+11 a c+4 b^2+2 b c-8 c^2)^2. \end{align}

Reference:

https://en.wikipedia.org/wiki/Positive_polynomial