<s>[INST] Prove that $x^2+xy^2+xyz^2 \ge 4xyz-4$ for positive real $x,y,z$ 
Prove that $x^2+xy^2+xyz^2 \ge 4xyz-4$ for postive real $x,y,z$.

I tried AM-GM but failed. I also can't apply Cauchy-Schwarz. I think we have to change this inequality to apply any well-known inequality. Please help me.
 [/INST] $$x^2+xy^2+xyz^2-4xyz+4 = x^2+xy^2-4xy+4+xy(z-2)^2 = $$ $$x^2+x(y-2)^2+xy(z-2)^2-4x+4  =(x-2)^2+x(y-2)^2+xy(z-2)^2\geq 0.$$
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