$$a^2 + b^2 + c^2 \geq ab + bc + ca,$$
so $k > 5$. Hence $k \ge 6$.

respectively, by the assumption in the problem, we have
$$k(1 \times 1 + 1 \times 2 + 1 \times 2) \le 5(1^2 + 1^2 + 2^2),$$
that is, $k \le 6$.
We will prove that $k = 6$ satisfies the requirement below. There is no harm in assuming $a \le b \le c$.
$$\begin{array}{lcl} \text{Since} & 6(ab + bc + ca) & > 5(a^2 + b^2 + c^2), \\[1em] \text{so} & 5c^2 - 6(a+b)c + 5a^2 + 5b^2 - 6ab & < 0, \\[1em] & \Delta = [6(a+b)]^2 - 4 \cdot 5(5a^2 + 5b^2 - 6ab) & \\ & = 64(ab - (a-b)^2) & \\ & \le 64ab \le 64 \cdot \frac{(a+b)^2}{4} & \\ & = 16(a+b)^2. & \\[1em] \text{Thus} & c < \frac{6(a+b) + \sqrt{\Delta}}{10} \le \frac{6(a+b) + 4(a+b)}{10} & \\ & = a+b. & \end{array}$$
Hence, there exists a triangle with $a, b$, and $c$ being the length of three sides.