Prove that for any set of three positive real $x, y, z$ such that $xy+yz+zx=3$
$x^2(y+z)+y^2(x+z)+z^2(x+y)+2\sqrt {xyz}\left(\sqrt{x^3+3x}+\sqrt{y^3+3x}+\sqrt{z^3+3x}\right)\ge$ $\ge 2xyz(x^2+y^2+z^2+6)$
Reasoning I know that the minimum value is with $x=y=z$ We have to find an equation equivalent to the date where we subtract and divide by xyz and don't appear other variables.
On
We need to prove that $$\sum_{cyc}\left(x^2y+x^2z+2\sqrt{xyz}\sqrt{x(x+y)(x+z)}\right)\geq2xyz(x+y+z)^2.$$ Now, by C-S $$\sqrt{x(x+y)(x+z)}\geq\sqrt{x}(x+\sqrt{yz})=\sqrt{x^3}+\sqrt{xyz}.$$ Thus, it's enough to prove that $$\sum_{cyc}(x^2y+x^2z+2\sqrt{x^4yz}+2xyz)\geq2xyz(x+y+z)^2$$ or $$\sum_{cyc}xy\sum_{cyc}(x^2y+x^2z+2\sqrt{x^4yz}+2xyz)\geq6xyz(x+y+z)^2$$ or $$\sum_{cyc}\left(x^3y^2+x^3z^2+2x^3yz+8x^2y^2z+2\sqrt{x^6y^3z}+2\sqrt{x^6z^3y}+2\sqrt{x^4y^3z^3}\right)\geq$$ $$\geq\sum_{cyc}(6x^3yz+12x^2y^2z)$$ or $$\sum_{cyc}\left(x^3y^2+x^3z^2-4x^3yz-4x^2y^2z+2\sqrt{x^6y^3z}+2\sqrt{x^6z^3y}+2\sqrt{x^4y^3z^3}\right)\geq0.$$ Now, $$\sum_{cyc}\left(\sqrt{x^6y^3z}+\sqrt{x^6z^3y}-2x^3yz\right)=\sum_{cyc}x^3\sqrt{yz}(\sqrt{y}-\sqrt{z})^2\geq0.$$ Thus, it remains to prove that $$\sum_{cyc}\left(x^3y^2+x^3z^2-4x^2y^2z+2\sqrt{x^4y^3z^3}\right)\geq0$$ or $$\sum_{cyc}z^2\left(x^3+y^3-2x^2y-2xy^2+2\sqrt{x^3y^3}\right)\geq0$$ or $$\sum_{cyc}\left(x^3-2\sqrt{x^3y^3}+y^3-2(x^2y+xy^2-2\sqrt{x^3y^3})\right)\geq0$$ or $$\sum_{cyc}z^2(\sqrt{x}-\sqrt{y})^2((x+\sqrt{xy}+y)^2-2xy)\geq0,$$ which is obvious.
Done!
We need to prove that $$\sum\limits_{cyc}xy\sum_{cyc}(x^2y+x^2z)+2\sum\limits_{cyc}xy\sum\limits_{cyc}x\sqrt{yz(x+y)(x+z)}\geq6xyz(x+y+z)^2.$$ Since by C-S $\sqrt{(x+y)(x+z)}\geq x+\sqrt{yz}$, it remains to prove that $$\sum\limits_{cyc}xy\sum_{cyc}(x^2y+x^2z)+2\sum\limits_{cyc}xy\sum\limits_{cyc}x\sqrt{yz}(x+\sqrt{yz})\geq6xyz(x+y+z)^2$$ or $$\sum\limits_{cyc}\left(x^3y^2+x^3z^2-4x^3yz-4x^2y^2z+2x^3y\sqrt{yz}+2x^3z\sqrt{yz}+2x^2\sqrt{y^3z^3}\right)\geq0$$ or $$\sum\limits_{cyc}\left(x^6y^4+x^6z^4-4x^6y^2z^2-4x^4y^4z^2+2x^6y^3z+2x^6z^3y+2x^4y^3z^3\right)\geq0$$ or $$\sum\limits_{cyc}\left(x^6y^4+x^6z^4-x^6y^3z-x^6z^3y\right)+2\sum\limits_{cyc}\left(x^6y^3z+x^6z^3y-2x^6y^2z^2\right)+$$ $$+\sum\limits_{cyc}\left(x^6y^3z+x^6z^3y-2x^4y^4z^2\right)-\sum\limits_{cyc}\left(2x^4y^4z^2-2x^4y^3z^3\right)\geq0.$$ Since $\sum\limits_{cyc}\left(x^6y^3z+x^6z^3y-2x^6y^2z^2\right)$ and $\sum\limits_{cyc}\left(x^6y^3z+x^6z^3y-2x^4y^4z^2\right)\geq0$,
it remains to prove that $$\sum\limits_{cyc}\left(x^6y^4+x^6z^4-x^6y^3z-x^6z^3y\right)+\sum\limits_{cyc}\left(x^6y^3z+x^6z^3y-2x^6y^2z^2\right)-$$ $$-\sum\limits_{cyc}\left(2x^4y^4z^2-2x^4y^3z^3\right)\geq0$$ or $$\sum\limits_{cyc}z^6(x-y)^2(x^2+xy+y^2)+\sum\limits_{cyc}z^6xy(x-y)^2-x^2y^2z^2\sum\limits_{cyc}z^2(x-y)^2\geq0$$ or $$\sum\limits_{cyc}z^4(x-y)^2(z^2(x+y)^2-x^2y^2)\geq0$$ or $$\sum\limits_{cyc}z^4(x-y)^2(xz+yz-xy)\geq0.$$ Let $x\geq y\geq z$.
Hence, $(x-z)^2\geq(x-y)^2$, $xy+yz-xz\geq0$ and $xy+xz-yz\geq0$.
Thus, $$\sum\limits_{cyc}z^4(x-y)^2(xz+yz-xy)\geq$$ $$\geq z^4(x-y)^2(xz+yz-xy)+y^4(x-z)^2(xy+yz-xz)\geq$$ $$\geq z^4(x-y)^2(xz-xy)+y^4(x-y)^2(xy-xz)=$$ $$=x(x-y)^2(y-z)^2(y+z)(y^2+z^2)\geq0$$ Done!