Denote the length of $AO$ by $x$, $BO$ by $y$ and $CO$ by $z$. From the cosine law we have: (fig.26)

Fig.26

$$AB = \sqrt{x^2 + xy + y^2}, \quad BC = \sqrt{y^2 + yz + z^2},$$
$CA = \sqrt{z^2 + zx + x^2}$ and so we can rewrite our inequality:
$$\frac{x^2}{\sqrt{y^2 + yz + z^2}} + \frac{y^2}{\sqrt{z^2 + zx + x^2}} + \frac{z^2}{\sqrt{x^2 + xy + y^2}} \ge \frac{x+y+z}{\sqrt{3}} \quad (1)$$

Without loss of generality we can suppose that $x \ge y \ge z$. Then it is easy to see that
$$\frac{1}{\sqrt{y^2 + yz + z^2}} \ge \frac{1}{\sqrt{x^2 + xz + z^2}} \ge \frac{1}{\sqrt{x^2 + xy + y^2}}. \text{ And also it is obvious that } x^2 \ge y^2 \ge z^2. \text{ Thus}$$
we can use an obvious inequality:
$$\frac{x^2}{\sqrt{y^2 + yz + z^2}} + \frac{y^2}{\sqrt{x^2 + xz + z^2}} + \frac{z^2}{\sqrt{x^2 + xy + y^2}} \ge \frac{y^2}{\sqrt{y^2 + yz + z^2}} + \frac{z^2}{\sqrt{x^2 + xz + z^2}} + \frac{z^2}{\sqrt{x^2 + xy + y^2}}, \\ \frac{x^2}{\sqrt{y^2 + yz + z^2}} + \frac{y^2}{\sqrt{x^2 + xz + z^2}} + \frac{z^2}{\sqrt{x^2 + xy + y^2}} \ge \frac{z^2}{\sqrt{y^2 + yz + z^2}} + \frac{x^2}{\sqrt{x^2 + xz + z^2}} + \frac{y^2}{\sqrt{x^2 + xy + y^2}}$$
Therefore the left hand side of the inequality (1) is no less than
$$\frac{1}{2} \left( \frac{y^2+z^2}{\sqrt{y^2+yz+z^2}} + \frac{x^2+z^2}{\sqrt{x^2+xz+z^2}} + \frac{z^2}{\sqrt{x^2+xy+y^2}} \right).$$
We will now show that $\frac{y^2+z^2}{\sqrt{y^2+yz+z^2}} \ge \frac{y+z}{\sqrt{3}}$. To prove this let us use the power mean inequality:
$$\frac{y^2+z^2}{\sqrt{y^2+yz+z^2}} \ge \sqrt{\frac{2}{3}(y^2+z^2)} = \frac{2}{\sqrt{3}} \sqrt{\frac{y^2+z^2}{2}} \ge \frac{2}{\sqrt{3}} \cdot \frac{y+z}{2} = \frac{y+z}{\sqrt{3}}$$
In the same way we can get analogous inequalities for pairs $(x, y)$ and $(x, z)$. And so we finally get:
$$\frac{y^2+z^2}{\sqrt{y^2+z^2}} + \frac{x^2+z^2}{\sqrt{x^2+xz+z^2}} + \frac{x^2+y^2}{\sqrt{x^2+xy+y^2}} \ge \frac{2(x+y+z)}{\sqrt{3}},$$
which was to be proved.

Here we will give another proof of the inequality (1). Using obvious inequalities $yz \le \frac{y^2+z^2}{2}$, $xz \le \frac{x^2+z^2}{2}$ and $xy \le \frac{x^2+y^2}{2}$ we can obtain:
$$\frac{x^2}{\sqrt{y^2 + yz + z^2}} + \frac{y^2}{\sqrt{z^2 + zx + x^2}} + \frac{z^2}{\sqrt{x^2 + xy + y^2}} \ge \sqrt{\frac{2}{3}} \left( \frac{x^2}{\sqrt{y^2 + z^2}} + \frac{y^2}{\sqrt{z^2 + y^2}} + \frac{z^2}{\sqrt{x^2 + y^2}} \right).$$
Now denote $S = x^2 + y^2 + z^2$ and consider the function $f(t) = \frac{1}{\sqrt{S-t}}$. Since
$$f''(t) = \frac{1}{\sqrt{(S-t)^3}} + \frac{3}{4} \frac{1}{\sqrt{(S-t)^5}} = \frac{4(S-t)+3t}{4\sqrt{(S-t)^5}} = \frac{4S-t}{4\sqrt{(S-t)^5}} \ge 0$$
for $0 \le t < S$, function $f$ is convex on $[0, S)$. And so applying the Jensen's inequality for $(0 \le x^2, y^2, z^2 < S$ we can get:
$$\frac{1}{3} \left( \frac{x^2}{\sqrt{y^2+z^2}} + \frac{y^2}{\sqrt{z^2+x^2}} + \frac{z^2}{\sqrt{x^2+y^2}} \right) = \frac{1}{3} \left( \frac{x^2}{\sqrt{S-x^2}} + \frac{y^2}{\sqrt{S-y^2}} + \frac{z^2}{\sqrt{S-z^2}} \right) = \\ = \frac{1}{3} \left( f(x^2) + f(y^2) + f(z^2) \right) \ge f\left(\frac{x^2+y^2+z^2}{3}\right) = \frac{8}{3\sqrt{S-\frac{8}{3}}} = \sqrt{\frac{(x^2+y^2+z^2)^2}{6}} \ge \frac{x+y+z}{3\sqrt{2}}$$
And finally
$$\frac{x^2}{\sqrt{y^2+yz+z^2}} + \frac{y^2}{\sqrt{z^2+zx+x^2}} + \frac{z^2}{\sqrt{x^2+xy+y^2}} \ge \sqrt{\frac{2}{3}} \cdot \frac{1}{\sqrt{2}} (x+y+z) = \frac{x+y+z}{\sqrt{3}},$$
And we're done.