Prove that $x^3+y^3+z^3+(x+y+z-1)^2\ge1+3xyz$ 
Let $x,y,z\ge0$ satisfy $\max\left \{ x,y,z \right \}\ge 1$. Prove that $$x^3+y^3+z^3+(x+y+z-1)^2\ge1+3xyz$$

My attempts:
From the condition we can deduce $x+y+z\ge 1$
The inequality can be written as $$(x+y+z)(x^2+y^2+z^2-xy-yz-zx)+(x+y+z)^2-2(x+y+z)\ge0$$ or $$x^2+y^2+z^2-xy-yz-zx+x+y+z-2\ge0$$
Let $x+y+z=p\ge1;xy+yz+zx=q$, the problem is: $$p^2-3q+p-2\ge0$$
I don't know how to find the relation between $p$ and $q$, because this inequality is not symmetric (The inequality hold iff $(x;y;z)=(1;0;0);(0;1;0);(0;0;1)$)
Please give me a hint in the comments, no need to give a full answer

Remark: I rewrote the proof using substitution $p = y + z, q = yz$.
WLOG, assume that $x \ge 1$. Let $p = y + z, q = yz$. We have $p^2 \ge 4q$.
We have
\begin{align*}
 &x^3 + y^3 + z^3 + (x + y + z - 1)^2 - 1 - 3xyz\\[5pt]
 =\, & x^3 + p^3 - 3pq + (x + p - 1)^2 - 1 - 3xq\\[5pt]
 =\,& x^3 + p^3 + (x + p - 1)^2 - 1 - 3(p + x)q \\[5pt]
 \ge\,& x^3 + p^3 + (x + p - 1)^2 - 1 - 3(p + x)\cdot \frac{p^2}{4}\\[5pt]
 =\,& (x + p)(x^2 - xp + p^2) + (x + p)^2 - 2(x + p) - \frac34(p + x)p^2\\[5pt]
 =\,& \frac14(x + p)[4x^2 - 4xp + 4p^2 + 4(x + p) - 8 - 3p^2]\\[5pt]
 =\,& \frac14(x + p)[p^2 - 4(x - 1)p + 4x^2 + 4x - 8]\\[5pt]
 =\,&\frac14(x + p)[(p - 2x + 2)^2 + 12(x - 1)]\\[5pt]
 \ge\,& 0.
\end{align*}
We are done.
