<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] $$x^2+xy^2+xyz^2+4=x^2+2\cdot\dfrac{xy^2}2+4\cdot\dfrac{xyz^2}4+4$$
Now using AM-GM inequality, $$\dfrac{x^2+2\cdot\dfrac{xy^2}2+4\cdot\dfrac{xyz^2}4+4}{1+2+4+1}\ge\sqrt[1+2+4+1]{x^2\cdot\left(\cdot\dfrac{xy^2}2\right)^2\cdot\left(\dfrac{xyz^2}4\right)^4\cdot4}$$
Here is how I've identified the coefficients:
let $$x^2+xy^2+xyz^2+4=a\cdot\dfrac{x^2}a+b\cdot\dfrac{xy^2}b+c\cdot\dfrac{xyz^2}c+d\cdot\dfrac4d$$
Now by AM-GM, $$\dfrac{a\cdot\dfrac{x^2}a+b\cdot\dfrac{xy^2}b+c\cdot\dfrac{xyz^2}c+d\cdot\dfrac4d}{a+b+c+d}\ge?$$
Compare the exponents of $x,y,z$ to find
$$a=d,b=2a,c=4a$$
Choose $a=1$
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