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

I have been trying to find a more aesthetic answer, but until I do, I will post a variational approach.
I have simplified the argument a bit, but I am still looking for a simpler approach.
To minimize $x^3+y^3+z^3-2x^2-2y^2-2z^2$ given $xyz=1$, we need to find $x,y,z$ so that for all variations $\delta x,\delta y,\delta z$ that maintain $xyz=1$; that is,
$$
\frac{\delta x}{x}+\frac{\delta y}{y}+\frac{\delta z}{z}=0\tag{1}
$$
we have $x^3+y^3+z^3-2x^2-2y^2-2z^2$ is stationary:
$$
(3x^2-4x)\,\delta x+(3y^2-4y)\,\delta y+(3z^2-4z)\,\delta z=0\tag{2}
$$
Standard orthogonality arguments imply that there is a $\lambda$ so that $x,y,z$ satisfy
$$
3t^2-4t=\frac\lambda{t}\implies3t^3-4t^2=\lambda\tag{3}
$$
Since $3t^3-4t^2$ decreases on $\left[0,\frac89\right]$ and increases for $t\gt\frac89$, for any value of $\lambda$, there can be at most two positive values for $x,y,z$; thus, two must be the same. Say $y=x$ and $z=x^{-2}$. Both $x$ and $x^{-2}$ must satisfy $(3)$, therefore,
$$
3x^3-4x^2=3x^{-6}-4x^{-4}\tag{4}
$$
Using Sturm's Theorem, we see that $3x^9-4x^8+4x^2-3$ has only one real root; that is $x=1$. Plugging this into the expression to be minimized gives a minimum of $-3$. This means
$$
x^3+y^3+z^3+3\ge2(x^2+y^2+z^2)\tag{5}
$$
