2. Find natural numbers $n$ such that for all positive numbers $a, b, c$, satisfying the inequality

$$
n(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)
$$

there exists a triangle with sides $a, b, c$.
2. Find natural numbers $n$ such that for all positive numbers $a, b, c$ satisfying the inequality

$$
n(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)
$$

there exists a triangle with sides $a, b, c$.

$\square$ Since $(a b+b c+c a) \leq\left(a^{2}+b^{2}+c^{2}\right)$, then $n>5$. If $n \geq 7$, then the numbers $a=1, b=1, c=2$ satisfy the inequality $n(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)$, but cannot be the lengths of the sides of a triangle. Therefore, it remains to check the value $n=6$. We will show that from the condition $6(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)$, where $0<a \leq b \leq c$, it follows that $c<a+b$ and therefore the numbers $a, b, c$ are the lengths of the sides of some triangle.

Indeed, considering the condition as a quadratic inequality in terms of $c$, we get

$$
5 c^{2}+6(a+b) c+\left[5\left(a^{2}+b^{2}\right)-6 a b\right]<0 ; \quad D / 4=16\left(a b-(a-b)^{2}\right) \leq 16 a b
$$

Therefore, taking into account that $2 \sqrt{a b} \leq a+b$, we conclude

$$
c<\frac{3(a+b)+4 \sqrt{a b}}{5} \leq a+b
$$