<s>[INST] 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
 [/INST] Continuing from OP's work. WLOG let $x \geq 1$, we have
$$q = xy+yz+zx = x(y+z) + yz \leq x(p-x) + (\frac{p-x}{2} ) ^2,$$
with equality iff $ y = z = \frac{ p-x}{2}$.
We WTS in the domain $ p \geq x \geq 1$,
$$p^2 + p  - 2 - \frac{1}{4} ( p-x)(p+3x) \geq 0.$$
Conditioning on each value of $p$, we have a much simpler quadratic in $x$:
$$ \frac{3}{4} ( x -p)(x+ \frac{p}{3}) + p^2 + p - 2 \geq 0.$$
By considering the quadratic on the restricted domain $ 1 \leq x \leq p$, the minimum occurs at $ \frac{ \frac{p}{3} + p} {2} = \frac{p}{3} $ if that's within the domain, else check the endpoints, giving us :

*

*Case 1 if $p \geq 3$: Minimum of $\frac{2p^2}{3} + p - 2 $ occurring at $x = \frac{p}{3}.  $

*Case 2 if $3 \geq p \geq 1$: Minimum of $ \frac{1}{4} (p-1)(3p+5)$ occurring at $ x = 1$.

Thus, for all $ p \geq 1$, the minimum at least 0.
Equality holds iff $ p = x = 1$, which gives us $ (x,y,z) =  (1, 0, 0)$ (and cyclic forms).
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