# Prove $\frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}} \geq \frac{\sqrt{(x+y)^2+(y+z)^2+(z+x)^2}}{2 \sqrt{3}}$ for $x,y,z \geq 0$

On the left we have arithmetic mean of pairwise quadratic means, which obeys:

$$\sqrt{\frac{x^2+y^2+z^2}{3}} \geq \color{blue}{ \frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}} } \geq \frac{x+y+z}{3}$$

On the right is quadratic mean of pairwise arithmetic means:

$$\sqrt{\frac{x^2+y^2+z^2}{3}} \geq \color{blue}{\frac{\sqrt{(x+y)^2+(y+z)^2+(z+x)^2}}{2 \sqrt{3}} } \geq \frac{x+y+z}{3}$$

By Wolfram Alpha it appears that:

$$\frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}} \geq \frac{\sqrt{(x+y)^2+(y+z)^2+(z+x)^2}}{2 \sqrt{3}}$$

How to prove it? I've started, but it's not working out so far.

First, we can transform the RHS into a more familiar form:

$$\frac{\sqrt{(x+y)^2+(y+z)^2+(z+x)^2}}{2 \sqrt{3}}=\sqrt{\frac{x^2+y^2+z^2+xy+yz+zx}{6}}$$

It makes sense that this expression is very close to arithmetic mean, because we have the following inequality:

$$\sqrt{\frac{x^2+y^2+z^2}{3}} \geq \frac{x+y+z}{3} \geq \sqrt{\frac{xy+yz+zx}{3}}$$

Let's try to prove the highlighted inequality. It is equivalent to:

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

Expanding the LHS we get:

$$2\sum_{cyc}\sqrt{x^2+y^2}\sqrt{y^2+z^2} \geq 3(xy+yz+zx)+x^2+y^2+z^2$$

To prove this inequality let's bound the sum on the LHS:

$$\sum_{cyc}\sqrt{x^2y^2+y^2z^2+z^2x^2+z^4} \geq 3\sqrt{x^2y^2+y^2z^2+z^2x^2} \geq \sqrt{3} (xy+yz+zx)$$

But this doesn't seem to prove the original inequality. So I'm not sure how to proceed.

You're almost done! $\sqrt{2(x^2+y^2)} \ge x+y$, and so $2\sqrt{x^2+y^2}\sqrt{x^2+z^2} \ge x^2+xy+xz+yz$. Summing gives the inequality you want.
• Note that the original inequality holds for all real $x,y,z$, not just non-negative ones, and the proof is the same. Jun 5, 2016 at 14:26