Example 6 Find all positive integers $k$ such that for any positive numbers $a, b, c$ 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 National Girls' Olympiad)
To find the positive integer $k$, we can first determine the upper and lower bounds of $k$, and then prove that the found $k$ satisfies the conditions.
From $a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a$, we know $k>5$. Taking $a, b, c$ that cannot form a triangle, such as $a=$ $b=1, c=2$, then $k(1 \cdot 1+1 \cdot 2+2 \cdot 1) \leqslant 5\left(1^{2}+1^{2}+2^{2}\right)$, we know $k \leqslant 6$. Therefore, $55\left(a^{2}+b^{2}+c^{2}\right)$. Then $a, b$, $c$ must form a triangle.
Assume $0<c$.
Construct the function $f(x)=5 x^{2}-6(a+b) x+5 a^{2}+5 b-6 a b$, then $f(c)f(c)$.
Assume $a+b \leqslant c$, then by $a+b, c \in\left[\frac{3}{5}(a+b)+\infty\right)$, we know $f(a+b) \leqslant f(c)$, which is a contradiction (since $f(x)$ is monotonically increasing to the right of the vertex).
Therefore, $a+b>c$, i.e., $a, b, c$ must form a triangle.