Minimium

Relevant wiki: Titu's Lemma

Since $a,b,c>0$ we can apply Titu's Lemma

$\dfrac{a^2}{1+2bc}+\dfrac{b^2}{1+2ac}+\dfrac{c^2}{1+2ab}\geq \dfrac{(a+b+c)^2}{3+2ab+2ac+2ab}$

Equality holds when $a=b=c=\dfrac{1}{\sqrt{3}}$ , substituting we get

$\begin{aligned} \dfrac{a^2}{1+2bc}+\dfrac{b^2}{1+2ac}+\dfrac{c^2}{1+2ab} & \geq \frac{\left(3\cdot\dfrac{1}{\sqrt{3}}\right)^2}{3+3\cdot\dfrac{2}{(\sqrt{3})^2}}\\ & \geq\dfrac{3}{5}=\boxed{0.6} \end{aligned}$

$\textbf{Note:}$

I'm assuming there is a typo in the text of the problem, since if you don't give the constraint that they are positive reals there is no global minimum