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

Here is a possible solution: (although it is not the most elegant one)
I will employ Mixing Variables technique here. Since the inequality is symmetric, WLOG let $x=\min(x,y,z)$. Therefore $t^2:=yz \ge 1$. 
Let 
$$f(x,y,z)=x^3+y^3+z^3-2(x^2+y^2+z^2)$$
I wish to show
$$f(x,y,z)\ge f(x,\sqrt{yz},\sqrt{yz}) = f(\frac1{t^2},t,t) \ge -3$$
Let us put $p^2=x, q^2=y, r^2=z$. The first inequality in the above chain is equivalent to
$$q^6+r^6-2q^3r^3 \ge 2(q^4+r^4-2q^2r^2)$$
$$\iff (q^3-r^3)^2 \ge 2(q^2-r^2)^2$$
$$\iff (q^2+qr+r^2)^2 \ge 2(q+r)^2$$
This is true since
$$(q^2+qr+r^2)^2 \ge q^4+r^4+2q^2r^2+2qr(q^2+r^2) \ge 4q^2r^2+2(q^2+r^2) \ge 2(q+r)^2$$
Therefore it enough to prove $f(\frac1{t^2},t,t)\ge -3$ for $t>0$ which is equivalent to
$$(t-1)^2((t^7-2t^5+t^3)+(t^7+t-2t^4)+(t^4-t^3+t^2)+t+1)\ge 0$$
Each term in the brackets of the large factor is greater than zero by AM-GM.
The last part is little tedious to do by hand. But you always know that $(t-1)$ has to be factor (possibly with multiplicity $2$) of that thing. That helps in simplification.
