Show that $ \frac{x^2}{\sqrt{x^2+y^2}} + \frac{y^2}{\sqrt{y^2+z^2}} + \frac{z^2}{\sqrt{z^2+x^2}} \geq \frac{1}{\sqrt{2}}(x+y+z) $ Show that for positive reals $x,y,z$ the following inequality holds and that the constant cannot be improved
$$
\frac{x^2}{\sqrt{x^2+y^2}} + \frac{y^2}{\sqrt{y^2+z^2}} + \frac{z^2}{\sqrt{z^2+x^2}} \geq \frac{1}{\sqrt{2}}(x+y+z)
$$
Background: I was digging through some old correspondence and found a letter from a very young me to Professor Love at the University of Melbourne. I had apparently ask via a letter (yes it was back when we wrote letters) how one could prove the above inequality (my version had $1/\sqrt{3}$ in it). He kindly wrote back but without a full proof. I just found the correspondence today and thought that this was a good question for this site.
Based on his letter and my old writings you can transform the above inequality as follows. First note that
$$
\text{g.l.b.}f(x,y,z) = \text{g.l.b.}f(x,z,y) = k \quad (say)
$$
where g.l.b is the greatest lower bound and $f(x,y,z)$ is the function
$$
\left(\frac{x^2}{\sqrt{x^2+y^2}} + \frac{y^2}{\sqrt{y^2+z^2}} + \frac{z^2}{\sqrt{z^2+x^2}}\right)\bigg/(x+y+z).
$$
and so
$$
\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2\phantom{y}} \geq 2k (x+y+z)
$$
Thus we need to prove that for positive reals $x,y$ and $x$ the following is true and tight:
$$
\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2\phantom{y}} \geq \sqrt{2} (x+y+z)
$$
One approach to prove this (used by Prof. Love) was to apply Hölder's inequality but this unfortunately only gives:
$$
\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2\phantom{y}} \geq \frac{2}{\sqrt{3}} (x+y+z)
$$
Geometric View: From a geometric view point this inequality can be viewed as stating that the perimeter of $\Delta PQR$ is not less than $\sqrt{2}$ times the sum of the the three edge-lengths of the box of sides $x,y,$ and $z$ and the points $P,Q$ and $R$ are three corners of the box that are not adjacent to each other.
I suspect that this is a "well known" inequality in the right circles but it is still not known to me. Thought that is was a nice problem for lovers of inequalities.

Another way.
By C-S and AM-GM we obtain:
$$\sum_{cyc}\frac{x^2}{\sqrt{x^2+y^2}}=\sqrt{\sum_{cyc}\left(\frac{x^4}{x^2+y^2}+\frac{2x^2y^2}{\sqrt{(x^2+y^2)(y^2+z^2)}}\right)}\geq$$
$$\geq\sqrt{\sum_{cyc}\frac{x^4}{x^2+y^2}+\frac{2(xy+xz+yz)^2}{\sum\limits_{cyc}\frac{x^2+y^2+y^2+z^2}{2}}}=\sqrt{\sum_{cyc}\frac{x^4}{x^2+y^2}+\frac{(xy+xz+yz)^2}{x^2+y^2+z^2}}$$ and it's enough to prove that:
$$\sum_{cyc}\frac{x^4}{x^2+y^2}+\frac{(xy+xz+yz)^2}{x^2+y^2+z^2}\geq\frac{(x+y+z)^2}{2}$$ or
$$\sum_{cyc}\left(\frac{x^4}{x^2+y^2}-\frac{3x^2-y^2}{4}\right)+\frac{(xy+xz+yz)^2}{x^2+y^2+z^2}\geq\frac{(x+y+z)^2}{2}-\frac{x^2+y^2+z^2}{2}$$ or
$$\sum_{cyc}\frac{(x^2-y^2)^2}{4(x^2+y^2)}\geq xy+xz+yz-\frac{(xy+xz+yz)^2}{x^2+y^2+z^2}$$ or
$$\sum_{cyc}(x-y)^2\left(\frac{(x+y)^2}{x^2+y^2}-\frac{2(xy+xz+yz)}{x^2+y^2+z^2}\right)\geq0$$ or
$$\sum_{cyc}\frac{(x-y)^2(x^2+y^2-xz-yz)^2}{x^2+y^2}\geq0.$$
