$\begin{array}{l}\text { II. ( } 50 \text { points) } \\ \text { Infinitely many non-degenerate } \triangle \text {, summing the shortest side, the second longest side, and the longest side of each } \\ \text { to form } 1 \text { triangle, what is the minimum value of the cosine of the largest angle of the resulting triangle? }\end{array}$
Let the shortest sides of an infinite number of triangles be $a_{1}, a_{2}, \cdots$, the middle sides be $b_{1}, b_{2}, \cdots$, and the longest sides be $c_{1}, c_{2}, \cdots$.
Where $a_{i}^{2}+b_{i}^{2} \geqslant c_{1}^{2}, a_{i} \leqslant b_{i} \leqslant c_{i}$.
Let $\mathrm{A}=\sum_{i=1}^{\infty} a_{i}, B=\sum_{i=1}^{\infty} b_{i}, C=\sum_{i=1}^{\infty} c_{\mathrm{r}}$
Then the cosine of the largest angle is $\frac{A^{2}+B^{2}-C^{2}}{2 A B}$.
We will prove that $\frac{A^{2}+B^{2}-C^{2}}{2 A B} \geqslant 1-\sqrt{2}$.
Since $a_{i} a_{j}+b_{i} b_{j}-\left(\sqrt{a_{i}^{2}+b_{i}^{2}} \cdot \sqrt{a_{j}^{2}+b_{j}^{2}}\right)$
$$
=a_{i} a_{j}+b_{i} b_{j}-\sqrt{\left(a_{i} a_{j}+b_{i} b_{j}\right)^{2}+\left(a_{i} b_{j}-b_{i} a_{j}\right)^{2}}
$$

And since $a_{i} a_{j}+b_{i} b_{j}>a_{i} b_{j}-b_{i} a_{j}$,
we have $\left(a_{i} a_{j}+b_{i} b_{j}\right)^{2}+\left(a_{i} b_{j}-b a_{j}\right)^{2}$
$$
\begin{array}{l}
\leqslant\left(a_{i} a_{j}+b_{i} b_{j}\right)^{2}+2(\sqrt{2}-1)\left(a_{i} a_{j}+b_{i} b_{j}\right)\left|a_{i} b_{j}-b_{i} a_{j}\right|+(\sqrt{2}-1)^{2}\left(a_{i} b_{j}\right. \\
\left.-b_{i} a_{j}\right)^{2}=a_{i} a_{j}+b_{i} b_{j}+(\sqrt{2}-1)\left|a_{i} b_{j}-b_{i} a_{j}\right|
\end{array}
$$

Therefore, $A^{2}+B^{2}-C^{2} \geqslant\left(\Sigma a_{i}\right)^{2}+\left(\Sigma b_{i}\right)^{2}-\left(\Sigma \sqrt{a_{i}^{2}+b_{i}^{2}}\right)^{2}$
$$
\begin{array}{l}
=2 \sum_{i \neq j} a_{i} a_{j}+2 \sum_{i \neq 1} b_{i} b_{j}-2 \sum_{i \neq j} \sqrt{\left(a_{i}^{2}+b_{i}^{2}\right)\left(a_{j}^{2}+b_{j}^{2}\right)} \\
\geqslant-(2 \sqrt{2}-2)\left|a_{i} b_{j}-b_{i} a_{j}\right|>-(2 \sqrt{2}-2)\left(\Sigma a_{i}\right)\left(\Sigma b_{i}\right)
\end{array}
$$

Therefore,
$$
\frac{A^{2}+B^{2}-C^{2}}{2 A B}>-\sqrt{2}+1
$$

Taking $a_{1}=a_{2}=\cdots=b_{2}=b_{3}=\cdots=a, c_{1}=b_{1}=b, c_{2}=c_{3}=\cdots=\sqrt{2} a, b$ much greater than $a$, at this point $\frac{A^{2}+B^{2}-C^{2}}{2 A B}$ approaches $-\sqrt{2}+1$.