Lower bound for $\frac{(x+y+z)^3}{xy+yz+zx}$ Let $x,y,z\geq 0$ and $x^2+y^2+z^2\geq 3$. What is the minimum value of $$D(x,y,z)=\frac{(x+y+z)^3}{xy+yz+zx}?$$
When $x=y=z=1$, $D(x,y,z)=9$. We have $(x+y+z)^2\leq 3(x^2+y^2+z^2)$ and $xy+yz+zx\leq x^2+y^2+z^2$, but these do not help directly with bounding $D(x,y,z)$.

Let $x^2+y^2+z^2=k(xy+xz+yz)$. 
Hence, by AM-GM we obtain:
$$\frac{(x+y+z)^3}{xy+xz+yz}\geq\frac{(x+y+z)^3}{xy+xz+yz}\sqrt{\frac{3}{x^2+y^2+z^2}}=$$
$$=\sqrt{\frac{3(x+y+z)^6}{(xy+xz+yz)^2(x^2+y^2+z^2)}}=\sqrt{\frac{3(k+2)^3}{k}}\geq\sqrt{\frac{3(3\sqrt[3]k)^3}{k}}=9.$$
The equality occurs for $x=y=z=1$, which says that the answer is $9$.
