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It should be noted, that the inequality is symmetrical, that is, the interchange in places of any two variables does not change its appearance. So, without any loss of generality we can assume that $a \ge b \ge c$. If we move all summands to one side and group them accordingly, the inequality becomes like this:
$$(\sqrt{a^2 + bc} - \sqrt{ca + ab}) + (\sqrt{b^2 + ca} - \sqrt{ab + bc}) + (\sqrt{c^2 + ab} - \sqrt{bc + ca}) \ge 0.$$
After multiplying every summand on the conjugate, we obtain that the following inequality should be proven:
$$\frac{(a-b)(a-c)}{\sqrt{a^2+bc}+\sqrt{ca+ab}} + \frac{(b-c)(b-a)}{\sqrt{b^2+ca}+\sqrt{ab+bc}} + \frac{(c-a)(c-b)}{\sqrt{c^2+ab}+\sqrt{bc+ca}} \ge 0.$$
Since $\frac{(a-b)(a-c)}{\sqrt{a^2+bc}+\sqrt{ca+ab}} \ge 0$ it will be enough to prove that:
$$\frac{(a-c)(b-c)}{\sqrt{c^2 + ab} + \sqrt{bc + ca}} \ge \frac{(b-c)(a-b)}{\sqrt{b^2 + ca} + \sqrt{ab + bc}}.$$
If $b-c=0$, hence there are zeros in both parts of the inequality. Otherwise, we can divide both parts by $b-c>0$. Further, since $a-c \ge b-c \ge 0$ it should be enough to prove the inequality:
$$\frac{1}{\sqrt{c^2 + ab} + \sqrt{bc + ca}} \ge \frac{1}{\sqrt{b^2 + ca} + \sqrt{ab + bc}} \Leftrightarrow \\ \sqrt{b^2 + ca} + \sqrt{ab + bc} \ge \sqrt{c^2 + ab} + \sqrt{bc + ca}.$$
The last inequality is true, since

$b^2 + ca \geq bc + ca$ and $ab + bc \geq c^2 + ab$.
$$$$
The inequality has been proven.