Prove inequality for positive real numbers $x,y,z$

Question

For positive, real numbers $x,y,z$ prove that
$$\frac{2(x+y+z)}{3}+\frac{3xyz}{xy+yz+zx}\ge3\sqrt[3]{xyz}$$
It's pretty obvious that the solution will have something to do with the $A.M-G.M$ inequality in my opinion, but I'm not sure how to go about this problem as everything I try leads me to a contradiction. For example, this is what I've tried…
$$x+y+z\ge3\sqrt[3]{xyz}$$
So we can prove that,
$$\frac{2(x+y+z)}{3}+\frac{3xyz}{xy+yz+zx}\ge x+y+z$$
$$\frac{3xyz}{xy+yz+zx}\ge \frac{x+y+z}{3}$$
$$9xyz\ge(x+y+z)(xy+yz+zx)$$
$$6xyz\ge x^2y+x^2z+y^2x+y^2z+z^2x+z^2y$$
But actually, from here, we can see that the opposite is actually true, so we obviously can't solve it this way. I've tried multiple other $A.M-G.M$ substitions but they all lead to similiar contradictions like this one. Can someone point me in the right direction and help me out? Thanks!

Answer 1

We have $$\begin{align} LHS&:=\frac{2(x+y+z)}{3}+\frac{3xyz}{xy+yz+zx}\\ &=\frac{(x+y+z)}{3}+\frac{(x+y+z)}{3}+\frac{3xyz}{xy+yz+zx}\\ &\ge 3\sqrt[3]{\frac{(x+y+z)^2xyz}{3(xy+yz+zx)}} \ge 3\sqrt[3]{\frac{3(xy+yz+zx)xyz}{3(xy+yz+zx)}}=3\sqrt[3]{xyz} \end{align}$$