<s>[INST] Prove the inequality $\displaystyle\sum_{s y m} x^{4} y^{2} z \geqslant 2 \sum_{s y m} x^{3} y^{2} z^{2}$ Prove for positive $x,y,z$ the inequality $x^{4} y^{2} z+x^{4} z^{2} y+y^{4} x^{2} z+y^{4} z^{2} x+z^{4} y^{2} x+z^{4} x^{2} y \geqslant 2(x^{3} y^{2} z^{2}+x^{2} y^{3} z^{2}+x^{2} y^{2} z^{3})$.
I found the following proof, it was without explanation:
\begin{gathered}
x^{3}(y+z)+y^{3}(x+x)+z^{3}(x+y) \geq 2(x^{2} y z+y^{2} z x+z^{2} x y) \\
2 x^{3}(y+z)+y^{3} z+z^{3} x \geq 6 x^{2} y z \\
2 y^{3}(z+x)+z^{3} x+x^{3} z \geq 6 y^{2} z x \\
2 z^{3}(x+y)+x^{3} y+y^{3} x \geq 6 z^{2} x y
\end{gathered}
I don't understand what starts happening from the 2nd line, can you explain and provide an alternative method, if possible?
 [/INST] The last term of the left side of the second line should be $z^3y$ and not $z^3x$. Its a typo. Also, they used the AM-GM inequality:
$2x^3(y+z) + y^3z + z^3y = x^3y+x^3y+x^3z+x^3z+y^3z+z^3y \ge 6\sqrt[6]{x^3y\cdot x^3y\cdot x^3z\cdot x^3z\cdot y^3z\cdot z^3y}=6x^2yz$
</s>