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?

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:
$2x^3(y+z) + y^3z + z^3y = x^3y+x^3y+x^3z+x^3z+y^3z+z^3y \ge 6\sqrt[6]{x^3y\cdot x^3y\cdot x^3z\cdot x^3z\cdot y^3z\cdot z^3y}=6x^2yz$
