Strange Equality Condition

Note that $\begin{aligned} x+y+z&=(x+y+z)\left (\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y} \right )\\ &= \dfrac{x(x+y+z)}{y+z}+\dfrac{y(y+z+x)}{z+x}+\dfrac{z(z+x+y)}{x+y}\\ &= \dfrac{x^2}{y+z}+\dfrac{(y+z)x}{y+z}+\dfrac{y^2}{z+x}+\dfrac{(z+x)y}{z+x}+\dfrac{z^2}{x+y}+\dfrac{(x+y)z}{x+y}\\ &= \dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}+(x+y+z)\\ \implies \dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y} &= 0 \end{aligned}$

Therefore, the expression is always 0.

Note that if $(x,y,z)$ is a solution to the equation $\frac{x}{y+z} + \frac{y}{x+z} + \frac{z}{x+y} = 1$ , then for any $r \in \mathbb{R}$ , $(rx,ry,rz)$ is also a solution. Let $Q(x,y,z)= \frac{x^2}{y+z} + \frac{y^2}{x+z} + \frac{z^2}{x+y}$ be the quantity to maximise. Note that $Q(rx,ry,rz) = r Q(x,y,z)$ . Hence there cannot be a maximum value unless $Q(x,y,z) =0$ .

$\text{Let} \; x+y+z = s \\ \text{Now} \sum_{cyclic} \dfrac{x^{2}}{y+z}= \sum_{cyclic} \dfrac{x^{2} -sx+sx}{s-x} = \sum_{cyclic} \left( \dfrac{x(x-s)}{s-x} + \dfrac{sx}{y+z} \right) = -x-y-z + s = -s+s =0$

$x + y + z = (x + y + z)(\dfrac{x}{y + z} + \dfrac{y}{x + z} + \dfrac{z}{x + y}) = \dfrac{x^2}{y + z} + \dfrac{xy}{x + z} + \dfrac{xz}{x + y}$ $+ \dfrac{xy}{y + z} + \dfrac{y^2}{x + z} + \dfrac{yz}{x + y} + \dfrac{xz}{y + z} + \dfrac{yz}{x + z} + \dfrac{z^2}{x + y} =$

$\dfrac{x^2}{y + z} + \dfrac{y^2}{x + z} + \dfrac{z^2}{x + y} + \dfrac{z(x + y)}{x + y} + \dfrac{x(y + z)}{y + z} + \dfrac{y(x + z)}{x + z} =$

$\dfrac{x^2}{y + z} + \dfrac{y^2}{x + z} + \dfrac{z^2}{x + y} + x + y + z \implies \boxed{\dfrac{x^2}{y + z} + \dfrac{y^2}{x + z} + \dfrac{z^2}{x + y} = 0}$