Problem 1174

Three, Proof: Without loss of generality, let $a+b+c=1$, then $a, b, c \in(0,1)$.
So $\frac{(b+c-a)^{2}}{(b+c)^{2}+a^{2}}=\frac{(1-2 a)^{2}}{(1-a)^{2}+a^{2}}$. We need to prove $\frac{(1-2 a)^{2}}{(1-a)^{2}+a^{2}} \geqslant \frac{23-54 a}{25}$.
$$\begin{array}{l} \frac{(1-2 a)^{2}}{(1-a)^{2}+a^{2}} \geqslant \frac{23-54 a}{25} \Leftrightarrow 25(1-2 a)^{2} \geqslant(23-54 a)\left(1-2 a+2 a^{2}\right) \\ \Leftrightarrow 108 a^{3}-54 a^{2}+2 \geqslant 0(*) \end{array}$$

Let $f(x)=108 x^{3}-54 x^{2}+2(x \in(0,1))$, then $f^{\prime}(x)=108 x(3 x-1)$.
It is known that when $x=\frac{1}{3}$, $f(x)$ reaches its minimum value, and $f\left(\frac{1}{3}\right)=0$.
Therefore, (*) holds, and thus $\frac{(1-2 a)^{2}}{(1-a)^{2}+a^{2}} \geqslant \frac{23-54 a}{25}$. Therefore,
$$\begin{array}{l} \quad \frac{(b+c-a)^{2}}{(b+c)^{2}+a^{2}}+\frac{(c+a-b)^{2}}{(c+a)^{2}+b^{2}}+\frac{(a+b-c)^{2}}{(a+b)^{2}+c^{2}}=\frac{(1-2 a)^{2}}{(1-a)^{2}+a^{2}}+\frac{(1-2 b)^{2}}{(1-b)^{2}+b^{2}}+\frac{(1-2 c)^{2}}{(1-c)^{2}+c^{2}} \geqslant \\ \frac{23-54 a}{25}+\frac{23-54 b}{25}+\frac{23-54 c}{25}=\frac{3}{5} . \end{array}$$