Inequality very very long

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Show that for any positive real numbers $x$, $y$, $z$ such that $xy + yz + zx = 3$, we have that: $$x^2(y+z)+y^2(z+x)+z^2(x+y)+2\sqrt{xyz}\left(\sqrt{x^3+3x}+\sqrt{y^3+3y}+\sqrt{z^3+3z}\right) \ge 2xyz\left(x^2+y^2+z^2+6\right)$$