II. (50 points)
Given that $x, y, z$ are all positive real numbers, find the minimum value of $\iota=\frac{(x+y-z)^{2}}{(x+y)^{2}+z^{2}}+\frac{(z+x-y)^{2}}{(z+x)^{2}+y^{2}}+\frac{(y+z-x)^{2}}{(y+z)^{2}+x^{2}}$.
When $x=y=z$, $t=\frac{3}{5}$.
Conjecture: $t \geqslant \frac{3}{5}$.
$$
\begin{array}{l} 
\geqslant \frac{3}{5} \Leftrightarrow \frac{z(x+y)}{(x+y)^{2}+z^{2}}+\frac{y(z+x)}{(z+x)^{2}+y^{2}}+\frac{x(y+z)}{(y+z)^{2}+x^{2}} \leqslant \frac{6}{5} . \\
\text { and } \frac{z(x+y)}{(x+y)^{2}+z^{2}} \leqslant \frac{2}{5}-\frac{9(x+y-2 z)}{25(x+y+z)} \\
\Leftrightarrow \frac{9(x+y-2 z)}{5(x+y+z)} \leqslant \frac{(x+y-2 z)[2(x+y)-z]}{(x+y)^{2}+z^{2}} \\
\Leftrightarrow 9(x+y-2 z)\left[(x+y)^{2}+z^{2}\right] \leqslant 5(x+y-2 z)[2(x+y)-z](x+y+z) \\
\Leftrightarrow(x+y-2 z)\left[10(x+y)^{2}+5 z(x+y)-5 z^{2}-9(x+y)^{2}-9 z^{2}\right] \geqslant 0 \\
\Leftrightarrow(x+y-2 z)\left[(x+y)^{2}+5 z(x+y)-14 z^{2}\right] \geqslant 0 \\
\Leftrightarrow(x+y-2 z)^{2}(x+y+7 z) \geqslant 0 \text { holds. }
\end{array}
$$

Similarly, $\frac{y(z+x)}{(z+x)^{2}+y^{2}} \leqslant \frac{2}{5}-\frac{9(z+x-2 y)}{25(x+y+z)}$,
$$
\frac{x(y+z)}{(y+z)^{2}+x^{2}} \leqslant \frac{2}{5}-\frac{9(y+z-2 x)}{25(x+y+z)},
$$

Adding inequalities (1), (2), and (3), we know that $t \geqslant \frac{3}{5}$. Therefore, the minimum value of $t$ is $\frac{3}{5}$, which can be achieved when $x=y=z$.