Question 14
It is known that
$$\frac a{b+c+d}+\frac b{c+d+a}+\frac c{d+a+b}+\frac d{a+b+c}=1.$$
Find
$$\frac {a^2}{b+c+d}+\frac {b^2}{c+d+a}+\frac {c^2}{d+a+b}+\frac {d^2}{a+b+c}.$$
$$\frac a{b+c+d}+\frac b{c+d+a}+\frac c{d+a+b}+\frac d{a+b+c}=1.$$
Find
$$\frac {a^2}{b+c+d}+\frac {b^2}{c+d+a}+\frac {c^2}{d+a+b}+\frac {d^2}{a+b+c}.$$