Ireland

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, $$\begin{gathered} \frac{a^2}{b+c+d}+\frac{b^2}{a+c+d}+\frac{c^2}{a+b+d} \\ \ge \frac{6a-b-c-d}{9}+\frac{6b-a-c-d}{9}+\frac{6c-a-b-d}{9} \\ =\frac{4a+4b+4c-3d}{9}. \end{gathered}$$ Equality occurs when $3a=b+c+d$, $3b=a+c+d$, $3c=a+b+d$ which leads to $a=b=c=d$.