1) Ta c/m BĐT sau:

Với a, b > 0 thì $a^3+b^3\ge ab\left(a+b\right)$

$\Leftrightarrow\left(a^3-a^2b\right)+\left(b^3-ab^2\right)\ge0$

$\Leftrightarrow a^2\left(a-b\right)+b^2\left(b-a\right)\ge0$

$\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0$ (luôn đúng vì a, b > 0)

Đẳng thức xảy ra $\Leftrightarrow a=b$

Như vậy ta có $\left\{{}\begin{matrix}x^3+y^3\ge xy\left(x+y\right)\\y^3+z^3\ge yz\left(y+z\right)\\z^3+x^3\ge zx\left(z+x\right)\end{matrix}\right.$

Do đó $VT\ge\dfrac{\sqrt{xyz+xy\left(x+y\right)}}{xy}+\dfrac{\sqrt{xyz+yz\left(y+z\right)}}{yz}+\dfrac{\sqrt{xyz+zx\left(z+x\right)}}{zx}$

$=\dfrac{\sqrt{xy\left(x+y+z\right)}}{xy}+\dfrac{\sqrt{yz\left(x+y+z\right)}}{yz}+\dfrac{\sqrt{zx\left(x+y+z\right)}}{zx}$

$=\sqrt{x+y+z}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)$

$=\sqrt{x+y+z}.\dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{xyz}}$

$=\sqrt{x+y+z}.\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)$

$\ge\sqrt{3\sqrt[3]{xyz}}.3\sqrt[3]{\sqrt{xyz}}=3\sqrt{3}$

Đẳng thức xảy ra $\Leftrightarrow x=y=z=1$