14. Let $k$ be a positive integer, such that for any positive numbers $a, b, c$ satisfying the inequality $k(a b+b c+c a)>5\left(a^{2}+b^{2}\right.$ $+c^{2}$), there must exist a triangle with side lengths $a, b, c$. Find the minimum value of $k$.


14. $k(a b+a c+b c)>5\left(a^{2}+b^{2}+c^{2}\right) \geqslant 5(a b+a c+b c), k \leqslant 5$ when, the inequality has no solution. When $k=6$, the inequality becomes $6(a b+a c+b c)>5\left(a^{2}+b^{2}+c^{2}\right)$. Without loss of generality, assume $c=\max (a, b, c)$, transform the inequality into $5 c^{2}-6(a+b) c+5\left(a^{2}+b^{2}\right)-6 a b0$, $5\left(a^{2}+b^{2}\right)-6 a b>0$, so the quadratic equation corresponding to the inequality has two positive roots $c_{1}, c_{2}$, the solution to the quadratic inequality in $c$ is $c \in\left(c_{1}, c_{2}\right)$. If $c_{2} \leqslant a+b$, then $c6$ when, $k(a b+a c+b c)>5\left(a^{2}+b^{2}+c^{2}\right)$ has solutions, but they cannot form a triangle, so the only $k$ that satisfies the condition is $k=6$.