from cyclic to symetric sums

Question

Can you explain to me how to pass from a cyclic to a symmetrical sum through weighed AM-GM (so I can use Muirhead's inequality) ? In particular by applying this to this inequality $$\sum_ {cyc} ^ {} a ^ 2b \ge \sum_ {cyc} ^ {} a ^ {\frac {5} {3}} b ^ {\frac {2} {3} } c ^ {\frac {2} {3}}$$

Answer 1

$$ \frac{a^2b + a^2b + c^2a}{3} \geq \sqrt[3]{a^5b^2c^2}. $$

Answer 2

For $x,y,z \geq 0$ \begin{eqnarray*} \sum_{cyc} x^3(xy+2z^2)(xy-z^2)^2 \geq 0 . \end{eqnarray*} Now expand, divide by $3$ rearrange \begin{eqnarray*} \sum_{cyc} x^6y^3 \geq \sum_{cyc} x^5y^2z^2 . \end{eqnarray*} Now substitue $a=x^3, b=y^3,c=z^3$ .