this inequality $\prod_{cyc} (x^2+x+1)\ge 9\sum_{cyc} xy$

Question

Let $x,y,z\in R$,and $x+y+z=3$

show that: $$(x^2+x+1)(y^2+y+1)(z^2+z+1)\ge 9(xy+yz+xz)$$

Things I have tried so far:$$9(xy+yz+xz)\le 3(x+y+z)^2=27$$ so it suffices to prove that $$(x^2+x+1)(y^2+y+1)(z^2+z+1)\ge 27$$ then the problem is solved. I stuck in here

Answer 1

Assume that $x,y,z$ are roots of the polynomial $$ q(w) = w^3-3w^2+s w-p \tag{1}$$ where $s=xy+xz+yz$ and $p=xyz$.

In order that $x,y,z\in\mathbb{R}$, the discriminant of $q$ must be non-negative, hence: $$ 54ps + 9s^2 \geq 4s^3+27p^2+108 p.\tag{2}$$ On the other hand, $2x+1,2y+1,2z+1$ are roots of the monic polynomial $8\cdot q\left(\frac{w-1}{2}\right)$, hence $(2x+1)^2,(2y+1)^2,(2z+1)^2$ are roots of the monic polynomial:

$$ w^3 + (8s-51) w^2 + (99-144 p+48 s+16 s^2) w + (-49 - 112 p - 64 p^2 - 56 s - 64 p s - 16 s^2) $$ and $4(x^2+x+1),4(y^2+y+1),4(z^2+z+1)$ are roots of the previous polynomial evaluated at $w-3$. Viete's theorem hence gives:

$$\prod_{cyc}(x^2+x+1) = 13-5 p+p^2+2 s+p s+s^2\tag{3}$$ and our problem boils down to proving that: $$ 13+p^2+p s+s^2 \geq 7s+5p.\tag{4} $$ However, that is trivial since the quadratic form $p^2+ps+s^2-7s-5p+13$ is positive definite and achieve its minimum, zero, at $(p,s)=(1,3)$.

Answer 2

with loss of generality, assume $x\ge y $

consider $f(x, y, z) = (x^2 + x + 1)(y^2 + y + 1) (z^2 + z + 1) - 9 (z(x+y) + xy)$

1. we first show that

$$f(x, y, z) \ge f((x+y)/2, (x+y)/2, z)$$

since $$f(x,y,z) - f((x+y)/2, (x+y)/2, z) = (x-y)((z^2 + z + 1)(z -2 - 2xy) + 9)$$

we simply show that

$$(z^2 + z + 1)(z -2 - 2xy) + 9 \ge 0$$

while $z^2 + z + 1 > 0$,

$$(z^2 + z + 1)(z - 2 - 2xy) + 9 \ge (z^2 + z + 1)(z - 2 - \frac{(3 - z)^2}{2}) + 9\\ = \frac{1}{2}(18 - (z^2 + z + 1)(z^2 - 8z +13))$$

and it is quite easy to show that when $0\le z\le 3$,

$$(z^2 + z + 1)(z^2 - 8z +13)\le 18$$ with maximum at $z = 1$.

2. If we achieve minimum of $f(x, y, z)$ at any point other than $(1,1,1)$, use the above argument, we can show a contradiction.

Answer 3

It's enough to prove that $$\prod_{cyc}\left(x^2+\frac{x(x+y+z)}{3}+\frac{(x+y+z)^2}{9}\right)-\frac{(xy+xz+yz)(x+y+z)^4}{9}\geq$$ $$\geq\frac{1}{2916}\left(\sum_{cyc}(5x^3+6x^2y+6x^2z-17xyz)\right)^2.$$ Let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v^2$ can be negative, and $xyz=w^3$.

Since $$\sum_{cyc}(5x^3+6x^2y+6x^2z-17xyz)=$$ $$=5(27u^3-27uv^2+3w^3)+6(9uv^2-3w^3)-51w^3=135u^3-81uv^2-54w^3$$ and $$54^2=2916,$$ we see that the last inequality is a linear inequality of $w^3$, which says that

it's enough to prove this inequality for an extremal value of $w^3$, which happens

for equality case of two variables.

Since the last inequality is even degree, we can assume $y=z=1$, which gives $$(x-1)^4(x+2)^2\geq0.$$ Done!