<s>[INST] How does $\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}$ relate to $\sqrt{x^2+y^2+z^2}$? Another possible 'mean' for three positive real numbers $x,y,z$ is made of pairwise quadratic means:
$$\frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}}$$
By QM-AM inequality it is greater than or equal to arithmetic mean of the three numbers:
$$\frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}} \geq \frac{x+y+z}{3}$$
Now what is its relationship to the quadratic mean of the three numbers:
$$\frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}} \text{ ? }  \sqrt{\frac{x^2+y^2+z^2}{3}}$$
Using the obvious inequality $x^2+y^2+z^2 \geq x^2+y^2$, we obtain the following relationship:
$$\sqrt{\frac{x^2+y^2+z^2}{3}} \geq  \sqrt{\frac{2}{3}} \frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}}   $$

Can we make this bound any tighter? How?

 [/INST] Never mind, it's easy to prove that
$$\sqrt{\frac{x^2+y^2+z^2}{3}} \geq   \frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}}$$
Square the RHS:
$$(\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2})^2=2(x^2+y^2+z^2)+2 \sum_{cyc} \sqrt{(x^2+y^2)(y^2+z^2)}$$
Using AM-GM we obtain:
$$\sqrt{(x^2+y^2)(y^2+z^2)} \leq \frac{x^2+2y^2+z^2}{2}$$
So it follows that:
$$(\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2})^2 \leq 6(x^2+y^2+z^2)$$
Or:
$$\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2} \leq \sqrt{6(x^2+y^2+z^2)}$$
Now we just divide by $3\sqrt{2}$.
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