This inequality seems to be true for all $x,y,z \geq 0$, but I'm not sure how to prove it:
$$(xy+yz+zx)^3 \geq xyz(x+y+z)^3$$
After expanding and simplifying we obtain:
$$x^3y^3+y^3z^3+z^3x^3 \geq xyz(x^3+y^3+z^3)$$
Which is the same as taking cube root of the original inequality with $a=x^3, b=y^3, c=z^3$:
$$ab+bc+ca \geq \sqrt[3]{abc}(a+b+c)$$
I initially thought to use Muirhead's inequality, but it doesn't work here, because the maximal exponent on the right is greater than on the left.
The best upper bound for the RHS I could find is:
$$xyz(x+y+z)^3 \leq (x^2+y^2+z^2)^{3/2}(xy+yz+zx)^{3/2}$$
But that's greater than the LHS as well:
$$(xy+yz+zx)^3 \leq (x^2+y^2+z^2)^{3/2}(xy+yz+zx)^{3/2}$$
So how to prove this?
It's wrong. Try $y=z=1$ and $x\rightarrow+\infty$.