How prove this $x^3+y^3+z^3+3\ge 2(x^2+y^2+z^2)$ Question:

let $x,y,z>0$ and such $xyz=1$, show that
  $$x^3+y^3+z^3+3\ge 2(x^2+y^2+z^2)$$

My idea: use AM-GM inequality
$$x^3+x^3+1\ge 3x^2$$
$$y^3+y^3+1\ge 3y^2$$
$$z^3+z^3+1\ge 3z^2$$
so
$$2(x^3+y^3+z^3)+3\ge 3(x^2+y^2+z^2)$$ 
But this is not my inequality,so How prove it? I know this condition is very important.but how use this condition? and this inequality is stronger

Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, our inequality is equivalent to $f(v^2)\geq0$, where $f$ is a linear function.
Hence, $f$ gets a minimal value, when $v^2$ gets an extremal value, which happens when
two numbers from $\{x,y,z\}$ are equal. 
Id est, it remains to prove our inequality for $y=x$ and $z=\frac{1}{x^2}$, which gives something obvious.
