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Knowing $x,y,z\ge0$ prove $x^2+xy^2+xyz^2\ge4xyz-4$

I thought that I should rearrange this inequality to be somewhat of the form of Schur's Inequality and WLOG I assumed $x\ge y\ge z$. Trying this way it did bring me out to nowhere so I tried to make some additions and subtractions of $+1,-1$ to factor but it didn't ring any bells.

So I tried this: $$x^2+xy^2+xyz^2\ge4xyz-4$$ $$x^2+xy^2+xyz^2-4xyz+4\ge0$$ $$x^2+xy^2+xyz^2-4xyz+4+4x-4x\ge0$$ $$(x+2)^2+xy^2+xyz^2-4xyz-4x\ge0$$ And while this did create some type of better structure for the problem it still didn't help me as I can't do anything about $xy^2$ or $xyz^2$. What I wanted to achieve with this method was some form of this : $$a(x+y)^2+b(y+z)^2+c(x+z)^2\ge0$$ Which is true as long as the condition set is $x,y,z\ge0$ I haven't tried using any famous inequality yet, but I do believe that I can make something out of this through $AM\ge GM$. Any help is appreciated.

Euclid
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    When you assume $x\ge y\ge z$ is not WLOG, because the inequality is not symmetrical in $x$, $y$ and $z$. – jjagmath May 27 '24 at 14:46

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Expanding on your idea, we can instead use $(x - 2)^2$, instead of $(x + 2)^2$. We get \begin{align*} 0 &\le (x - 2)^2 + xy^2 + xyz^2 - 4xyz + 4x \\ &= (x - 2)^2 + x(y^2 - 4y + 4 + yz^2 - 4yz + 4y) \\ &= (x - 2)^2 + x((y - 2)^2 + y(z^2 - 4z + 4)) \\ &= (x - 2)^2 + x((y - 2)^2 + y(z - 2)^2), \end{align*} which is true, and equivalent to the original inequality.

Theo Bendit
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