Ukrainian National Mathematical Olympiad

Multiplying the given equality by $(a+b+c+d)$, we obtain $$\begin{array}{rl}& \frac{a(a+b+c+d)}{b+c+d}+\frac{b(a+b+c+d)}{c+d+a}+\frac{c(a+b+c+d)}{d+a+b}+\frac{d(a+b+c+d)}{a+b+c}=\\ & =\frac{a^2+a(b+c+d)}{b+c+d}+\frac{b^2+b(c+d+a)}{c+d+a}+\frac{c^2+c(d+a+b)}{d+a+b}+\frac{d^2+d(a+b+c)}{a+b+c}=\\ & =\frac{a^2}{b+c+d}+a+\frac{b^2}{c+d+a}+b+\frac{c^2}{d+a+b}+c+\frac{d^2}{a+b+c}+d=a+b+c+d.\end{array}$$ It then follows that $$\frac{a^2}{b+c+d}+\frac{b^2}{c+d+a}+\frac{c^2}{d+a+b}+\frac{d^2}{a+b+c}=0.$$