Example 3 Try to find all positive integers $k$, such that for any positive integers $a, b, c$ satisfying the inequality
$$
k(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)
$$

there must exist a triangle with side lengths $a$, $b$, and $c$.
(2002, Girls' Mathematical Olympiad)
Explanation: First, from the inequality relationship between $a b+b c+c a$ and $a^{2}+b^{2}+c^{2}$, we derive $k>5$. Then, by constructing an example, we find that $k \leqslant 6$. Therefore, since $k \in \mathbf{Z}_{+}$, we conclude that $k=6$. Finally, we provide the proof.

Notice that for any positive real numbers $a, b, c$, we have
$$
a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a.
$$

On one hand, from
$$
\begin{array}{l}
k(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right) \\
\geqslant 5(a b+b c+c a),
\end{array}
$$

we get $k>5$.
Since $k \in \mathbf{Z}_{+}$, we have $k \geqslant 6$.
On the other hand, because a triangle with side lengths $1, 1, 2$ does not exist, by the problem statement, we have
$$
k(1 \times 1+1 \times 2+2 \times 1) \leqslant 5\left(1^{2}+1^{2}+2^{2}\right),
$$

which implies $k \leqslant 6$.
Thus, we have $k=6$.
Next, we prove that $k=6$ satisfies the condition.
Assume without loss of generality that $a \leqslant b \leqslant c$.
From $6(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)$, we get
$5 c^{2}-6(a+b) c+\left(5 a^{2}+5 b^{2}-6 a b\right)<0$.
By $\Delta=64\left[a b-(a-b)^{2}\right]$
$$
\leqslant 64 a b \leqslant 16(a+b)^{2},
$$

we have $c<\frac{6(a+b)+\sqrt{\Delta}}{10}$
$$
\leqslant \frac{6(a+b)+4(a+b)}{10}=a+b.
$$

Therefore, a triangle with side lengths $a, b, c$ exists.
Hence, the positive integer $k$ that satisfies the condition is $k=6$.