35. 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 $a, b, c$. (2002 Year
35. From $(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geqslant 0$, we get $a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a$, so, $k>5$, hence $k \geqslant 6$.

Since there does not exist a triangle with side lengths $1,1,2$, according to the problem, we have $k(1 \times 1+1 \times 2+1 \times 2) \leqslant 5\left(1^{2}+1^{2}+2^{2}\right)$, i.e., $k \leqslant 6$. $\square$

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 $5 c^{2}-6(a+b) c+5 a^{2}+5 b^{2}-6 a b<0, \Delta=[6(a+b)]^{2}-4 \times 5\left(5 a^{2}+5 b^{2}-6 a b\right)=64[a b-(a-$ $\left.b)^{2}\right] \leqslant 64 a b \leqslant 64 \times \frac{(a+b) 2}{4}=16(a+b)^{2}$, thus
$$c<\frac{6(a+b)+\sqrt{\Delta}}{10} \leqslant \frac{6(a+b)+4(a+b)}{10}=a+b$$

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