Problem. Let $x,y,z\ge 0: x^2+y^2+z^2+xyz=4.$ Prove that$$4(xy+yz+zx-xyz) \geqslant (x^2y+z)(y^2z+x)(z^2x+y).$$ It was here.
The condition implies me set the substitution $$x=\frac{2a}{\sqrt{(a+b)(a+c)}};y=\frac{2b}{\sqrt{(a+b)(b+c)}};z=\frac{2c}{\sqrt{(c+b)(a+c)}}.$$ The rest is quite complicated. I think we can find a better ideas. Thank you very much.
Some thoughts.
We use pqr method.
Let $p = x + y + z, q = xy + yz + zx, r = xyz$.
The condition $x^2+y^2+z^2+xyz=4$ becomes $$p^2 - 2q + r = 4. \tag{1}$$
From (1), we have $$0 \le r \le 1, \quad 2 \le p \le 3, \quad q = \frac12 p^2 + \frac12 r - 2. \tag{2}$$
We have \begin{align*} &(x^2y + z)(y^2z + x)(z^2x + y)\\ ={}&{x}^{3}{y}^{3}{z}^{3}+{x}^{4}y{z}^{2}+{x}^{2}{y}^{4}z+x{y}^{2}{z}^{4}+ {x}^{3}{y}^{2}+{z}^{3}{x}^{2}+{y}^{3}{z}^{2}+xyz\\ ={}& r^3 + r(xy^3 + yz^3 + zx^3) + (x^3y^2 + y^3z^2 + z^3x^2) + r\\ ={}& (x^2y + y^2z + z^2x) (q - pr) +{p}^{2}qr-{p}^{2}r-2\,p{r}^{2}-{q}^{2}r+{r}^{3 }+rq+r\\ \le{}& \left(\frac{4}{27}p^3 - r\right) (q - pr) +{p}^{2}qr-{p}^{2}r-2\,p{r}^{2}-{q}^{2}r+{r}^{3 }+rq+r \tag{3} \end{align*} where we use $q - pr \ge \sqrt{3pr} - pr \ge \sqrt{3p}\, r - pr \ge 0$ (using $q^2 \ge 3pr$ and $r \le 1$ and $p\le 3$), and the known inequality $x^2y + y^2z + z^2x \le \frac{4}{27}(x + y + z)^3 - xyz$.
It suffices to prove that $$4(q - r) \ge \left(\frac{4}{27}p^3 - r\right) (q - pr) +{p}^{2}qr-{p}^{2}r-2\,p{r}^{2}-{q}^{2}r+{r}^{3 }+rq+r. \tag{4}$$
From (2) and (4), using $q = \frac12 p^2 + \frac12 r - 2$, it suffices to prove that \begin{align*} g(p, r) &:= -81\,{r}^{3}+ \left( 108\,p-216 \right) {r}^{2}+ \left( -11\,{p}^{4}-8 \,{p}^{3}+108\,{p}^{2}+108 \right) r\\ &\qquad + 8(p-2)(3-p)(p+2)(p^2 + 3p + 9) \ge 0.\tag{5} \end{align*}
From (2), using $q = \frac12 p^2 + \frac12 r - 2$, we have \begin{align*} 0 &\le (x - y)^2(y - z)^2(z - x)^2\\ &= -4\,{p}^{3}r+{p}^{2}{q}^{2}+18\,pqr-4\,{q}^{3}-27\,{r}^{2}\\ &= \frac{p^2 + 4p + 2r + 4}{4}[-{r}^{2}+ \left( -2\,{p}^{2}+20\,p-40 \right) r-{p}^{4}+4\,{p}^{3}+4\, {p}^{2}-32\,p+32 ] \end{align*} which results in $$f(p, r) := -{r}^{2}+ \left( -2\,{p}^{2}+20\,p-40 \right) r-{p}^{4}+4\,{p}^{3}+4\, {p}^{2}-32\,p+32 \ge 0. \tag{6}$$
We have \begin{align*} &g(p, r) - 81r \cdot f(p, r) \\ ={}& \left( 162\,{p}^{2}-1512\,p+3024 \right) {r}^{2}+ \left( 70\,{p}^{4}- 332\,{p}^{3}-216\,{p}^{2}+2592\,p-2484 \right) r\\ &\qquad + 8(p-2)(3-p)(p+2)(p^2 + 3p + 9)\\ ={}& 8( p-2)( 3-p ) ( p+2 ) ( {p}^{2}+3\,p+9) ( 1-r ) ^{2}\\ &\qquad + ( -16\,{ p}^{5}+70\,{p}^{4}-268\,{p}^{3}+216\,{p}^{2}+2592\,p-4212) r ( 1-r ) \\ &\qquad + ( -8\,{p}^{5}+70\,{p}^{4}-300\,{p}^{3}+162 \,{p}^{2}+1080\,p-324) {r}^{2}\\ \ge{}& 0 \tag{7} \end{align*} where we use $ -16\,{ p}^{5}+70\,{p}^{4}-268\,{p}^{3}+216\,{p}^{2}+2592\,p-4212\ge 0$ and $-8\,{p}^{5}+70\,{p}^{4}-300\,{p}^{3}+162 \,{p}^{2}+1080\,p-324 \ge 0$ for all $2\le p \le 3$ (easy to prove), and $0 \le r \le 1$.
From (5), (6) and (7), the desired result follows.
We are done.
I hope the following will help.
Let $x=a-2$, $y=b-2$ and $z=c-2$, where $\{a,b,c\}\subset[2,4].$
Thus, the condition gives $$\sum_{cyc}(2ab-a^2)=abc$$ and we need to prove that $$4\left(\sum_{cyc}\left(a-\tfrac{2abc}{\sum\limits_{cyc}(2ab-a^2)}\right)\left(b-\tfrac{2abc}{\sum\limits_{cyc}(2ab-a^2)}\right)\cdot\tfrac{abc}{\sum\limits_{cyc}(2ab-a^2)}-\prod_{cyc}\left(a-\tfrac{2abc}{\sum\limits_{cyc}(2ab-a^2)}\right)\right)\geq$$ $$\geq\prod_{cyc}\left(\left(a-\tfrac{2abc}{\sum\limits_{cyc}(2ab-a^2)}\right)^2\left(b-\tfrac{2abc}{\sum\limits_{cyc}(2ab-a^2)}\right)+\left(c-\tfrac{2abc}{\sum\limits_{cyc}(2ab-a^2)}\right)\cdot\tfrac{a^2b^2c^2}{\left(\sum\limits_{cyc}(2ab-a^2)\right)^2}\right)$$ or $$4a^4b^4c^4\left(\sum_{cyc}ab(2ab+2ac-a^2-b^2-c^2)(2ab+2bc-a^2-b^2-c^2)-\prod_{cyc}(2ab+2ac-a^2-b^2-c^2)\right)$$ $$\geq\prod_{cyc}\left((2ab+2ac-a^2-b^2-c^2)^2(2ab+2bc-a^2-b^2-c^2)+c^3b(2ac+2bc-a^2-b^2-c^2)\right),$$ which is $18$-th degree homogeneous polynomial inequality.
Also, the condition $\{a,b,c\}\subset[2,4]$ gives $$a+b-c\geq2+2-4\geq0$$ and we can assume that $a=v+w$, $b=u+w$ and $c=u+v,$ where $u$, $v$ and $w$ are non-negatives.
Also, $$2ab+2ac-a^2-b^2-c^2=$$ $$=2(w^2+v^2+2(uv+uw+vw)-u^2-v^2-w^2-uv-uw-vw)=$$ $$=2(uv+uw+vw-u^2)\geq0,$$ $$uv+uw+vw-v^2\geq0$$ and $$uv+uw+vw-w^2\geq0.$$
