3 Find all positive integers $k$ such that for any positive numbers $a$, $b$, and $c$ satisfying the inequality $k(ab + bc + ca) > 5(a^2 + b^2 + c^2)$, there must exist a triangle with side lengths $a$, $b$, and $c$.
(Qian Zhanwang)
From $(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geqslant 0$, we get
$$
a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a,
$$

Therefore, $k>5$, so $k \geqslant 6$.
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 $k \leqslant 6$.

Next, we prove that $k=6$ satisfies the requirement. Without loss of generality, assume $a \leqslant b \leqslant c$,
since
$$
6(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right),
$$

we have
$$
\begin{array}{l} 
5 c^{2}-6(a+b) c+5 a^{2}+5 b^{2}-6 a b<0 \\
\Delta=[6(a+b)]^{2}-4 \cdot 5\left(5 a^{2}+5 b^{2}-6 a b\right) \\
=64\left[a b-(a-b)^{2}\right] \\
\leqslant 64 a b \leqslant 64 \cdot \frac{(a+b)^{2}}{4} \\
=16(a+b)^{2}, \\
c<\frac{6(a+b)+\sqrt{\Delta}}{10} \leqslant \frac{6(a+b)+4(a+b)}{10} \\
=a+b,
\end{array}
$$

Thus,
$$
\begin{aligned}
c & <\frac{6(a+b)+\sqrt{\Delta}}{10} \leqslant \frac{6(a+b)+4(a+b)}{10} \\
& =a+b,
\end{aligned}
$$

Therefore, there exists a triangle with side lengths $a, b, c$.