# How does $(\sqrt{xy}+\sqrt{yz}+\sqrt{zx})/3$ relate to $\sqrt[3]{xyz}$ for $x,y,z>0$?

This kind of 'mean' for three positive real numbers $x,y,z$ appears in some applications (elliptic integrals for example) and various inequality problems.

We have, by AM-GM-HM inequalities for pairs of numbers:

$$\frac{x+y+z}{3} \geq \frac{\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}{3} \geq \frac{2}{3} \left(\frac{xy}{x+y}+\frac{yz}{y+z}+\frac{zx}{z+x} \right)$$

And for three numbers we have:

$$\frac{x+y+z}{3} \geq \sqrt[3]{xyz} \geq \frac{3xyz}{xy+yz+zx}$$

But can we relate this expression to the geometric mean for three numbers (for arbitrary $x,y,z>0$):

$$\frac{\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}{3} ~?~ \sqrt[3]{xyz}$$

Edit

In hindsight, that's so simple, it's not even worth asking. As Jez said, we apply AM-GM and get:

$$\frac{\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}{3} \geq \sqrt[3]{xyz}$$

Just apply AM-GM to $\sqrt{xy}$, $\sqrt{yz}$ and $\sqrt{zx}$.