Three, try to 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 $a, b, c$.
Three, due to $(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geqslant 0$,
thus $a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a$.
It can be known that $k>5$. Noting that $k$ is a positive integer, therefore, $k \geqslant 6$.
Since there does not exist a triangle with side lengths $1,1,2$, according to the problem, we have
$$
k(1 \times 1+1 \times 2+1 \times 2) \leqslant 5\left(1^{2}+1^{2}+2^{2}\right),
$$

which means $k \leqslant 6$.
The following proves that $k=6$ satisfies the problem's requirements.
Assume without loss of generality that $a \leqslant b \leqslant c$, then
$$
6(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right),
$$

that is
$$
\begin{array}{l}
5 c^{2}-6(a+b) c+5 a^{2}+5 b^{2}-6 a b5\left(a^{2}+b^{2}+c^{2}\right)
\end{array}
$$
a contradiction. Hence $c<a+b$.
Method two: Construct a function
$$
f(x)=5 x^{2}-6(a+b) x+5 a^{2}+5 b^{2}-6 a b \text {. }
$$

Then $f(c)<0$.
Since $f(x)$ is increasing in the interval $\left[\frac{3}{5}(a+b),+\infty\right)$, and
$$
\begin{aligned}
f(a+b)= & 5(a+b)^{2}-6(a+b)(a+b)+5 a^{2}+ \\
& 5 b^{2}-6 a b=4(a-b)^{2} \geqslant 0,
\end{aligned}
$$

thus $c<a+b$.