Example 5 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 $a, b, c$.
(First China Girls Mathematical Olympiad)
Analysis: To find $k$, we can first determine the upper and lower bounds of $k$, and then argue that the $k$ found satisfies the conditions of the problem.
Solution: It is easy to know that $a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a$.
Therefore, the required $k>5$.
Take $a, b, c$ that do not form a triangle ($a=1, b=1, c=2$), then we have
$$
k(1 \times 1+1 \times 2+2 \times 1) \leqslant 5\left(1^{2}+1^{2}+2^{2}\right).
$$

This gives $k \leqslant 6$.
Therefore, if $55\left(a^{2}+b^{2}+c^{2}\right),
$$

then $a, b, c$ must form a triangle.
Assume without loss of generality that $0c$.
Construct the function
$$
f(x)=5 x^{2}-6(a+b) x+5 a^{2}+5 b^{2}-6 a b,
$$

then $f(c)f(c)$.
Assume $a+b \leqslant c$, by
$$
a+b, c \in\left[\frac{3}{5}(a+b),+\infty\right)
$$

we know $f(a+b) \leqslant f(c)$.
Contradiction (since $f(x)$ is monotonically increasing to the right of the vertex).
Therefore, $a+b>c$, i.e., $a, b, c$ form a triangle.
Note: To compare the sizes of $a+b$ and $c$, one can first compare the sizes of $f(a+b)$ and $f(c)$, and then deduce the sizes of $a+b$ and $c$ based on the monotonicity of $f(x)$. Of course, this is predicated on constructing an appropriate function $f(x)$.