For arbitrary real numbers $x$, $y$ we have $9x^2 - 6xy + y^2 = (3x - y)^2 \ge 0$. Hence, if $y > 0$, $\frac{x^2}{y} \ge \frac{6x-y}{9}$ with equality iff $y = 3x$. Hence,
$$\frac{a^2}{b+c+d} + \frac{b^2}{a+c+d} + \frac{c^2}{a+b+d} \\ \ge \frac{6a-b-c-d}{9} + \frac{6b-a-c-d}{9} + \frac{6c-a-b-d}{9} \\ = \frac{4a+4b+4c-3d}{9}.$$
Equality occurs when $3a = b + c + d$, $3b = a + c + d$, $3c = a + b + d$ which leads to $a = b = c = d$.