A strengthened inequality of tianty

Problem. (Ji Chen) If $ x,y,z$ be nonnegative real numbers such that $ x+y+z=1$, then $$ \prod \left[(y - z)^2 + 3\sqrt {3}yz\right]\geqslant 3\sqrt{3} xyz(x+y+z)^3$$

Solution. (tthnew) \begin{align*} \displaystyle \text{LHS}-\text{RHS} =& \,\left( 26-15\,\sqrt {3} \right) \left( xy+zx+yz \right) \sum yz \left( x- y \right) \left( x-z \right)\\& +\left( x+y+z \right) \sum yz \left( y+z \right) \left( x-y \right) \left( x-z \right)\\& +3 \left(6\sqrt {3} -10 \right) {y}^{2}{z}^{2} \left( x-y \right) \left( x-z \right) \geqslant 0 \end{align*}