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

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!

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}