Question 147
If $x$, $y$, $z$ and $w$ are real numbers such that
$$
\frac{x}{y+z+w} + \frac{y}{z+w+x} + \frac{z}{w+x+y} + \frac{w}{x+y+z} = 1,
$$
find
$$
\frac{x^2}{y+z+w} + \frac{y^2}{z+w+x} + \frac{z^2}{w+x+y} + \frac{w^2}{x+y+z}.
$$
$$
\frac{x}{y+z+w} + \frac{y}{z+w+x} + \frac{z}{w+x+y} + \frac{w}{x+y+z} = 1,
$$
find
$$
\frac{x^2}{y+z+w} + \frac{y^2}{z+w+x} + \frac{z^2}{w+x+y} + \frac{w^2}{x+y+z}.
$$