<s>[INST] Prove $(x+y)(y+z)(z+x)\ge\frac{8}{3}(x+y+z)\sqrt[3]{x^2y^2z^2}$ 
If $x$, $y$ and $z$ are positive numbers,prove:
  $$(x+y)(y+z)(z+x)\ge\frac{8}{3}(x+y+z)\sqrt[3]{x^2y^2z^2}.$$
Additional info:I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM because I have background in them.

Things I have done so far: We know that $$(x+y)(y+z)(z+x)\ge8xyz$$$$\frac{8}{3}(x+y+z)\sqrt[3]{x^2y^2z^2}\ge8xyz$$
So $(x+y)(y+z)(z+x)\ge\frac{8}{3}(x+y+z)\sqrt[3]{x^2y^2z^2}$ is stronger inequality than  $(x+y)(y+z)(z+x)\ge8xyz$. My Problem in Proving this inequality is that $\frac{1}{3}$ and $\sqrt[3]{}$ makes me think applying $3$ term AM-GM.but there is $8$ terms on Left hand side.Also I can't think of way to make $8$ on right hand side by $3$ term AM-GM.
By writing LHS in expanded form I was able to reach this$$LHS+xyz \ge 3\sum_{cyc}x\sqrt[3]{x^2y^2z^2}$$
 [/INST] Well I think i figured it out.by following what I have written in my post:$$LHS\ge 3(\sum_{cyc}x)\sqrt[3]{x^2y^2z^2} - xyz \ge \frac{8}{3}(x+y+z)\sqrt[3]{x^2y^2z^2}$$
For Proving $3(\sum_{cyc}x)\sqrt[3]{x^2y^2z^2} - xyz \ge \frac{8}{3}(x+y+z)\sqrt[3]{x^2y^2z^2}$ :$$\sqrt[3]{x^2y^2z^2}(3(\sum_{cyc}x) - \sqrt[3]{xyz})\ge\frac{8}{3}(x+y+z)\sqrt[3]{x^2y^2z^2}$$ $$3(\sum_{cyc}x) - \sqrt[3]{xyz} \ge \frac{8}{3}(x+y+z)$$ $$\frac{x+y+z}{3}\ge\sqrt[3]{xyz}$$
Which is true by AM-GM.
</s>