China Girls' Mathematical Olympiad

$$a^2+b^2+c^2\ge ab+bc+ca,$$ 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 $a\le b\le c$. $$\begin{array}{lcl}\text{Since}& 6(ab+bc+ca)& >5(a^2+b^2+c^2),\\ \text{so}& 5c^2-6(a+b)c+5a^2+5b^2-6ab& <0,\\ & \Delta =[6(a+b){]}^2-4\cdot 5(5a^2+5b^2-6ab)& \\ & =64(ab-(a-b{)}^2)& \\ & \le 64ab\le 64\cdot \frac{(a+b{)}^2}{4}& \\ & =16(a+b{)}^2.& \\ \text{Thus}& c<\frac{6(a+b)+\sqrt{\Delta }}{10}\le \frac{6(a+b)+4(a+b)}{10}& \\ & =a+b.& \end{array}$$ Hence, there exists a triangle with $a,b$, and $c$ being the length of three sides.