How does $\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}$ relate to $\sqrt{x^2+y^2+z^2}$?

Question

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?

Answer 1

If $\sqrt{x^2+y^2}=c$, $\sqrt{x^2+z^2}=b$ and $\sqrt{y^2+z^2}=a$ so we can use $\sqrt{3(a^2+b^2+c^2)}\geq a+b+c$.

Answer 2

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 $3\sqrt{2}$.