5 Find all real numbers $a$ such that any positive integer solution of the inequality $x^{2}+y^{2}+z^{2} \leqslant a(x y+y z+z x)$ are the lengths of the sides of some triangle.
5. Taking $x=2, y=z=1$, we have $a \geqslant \frac{6}{5}$. Therefore, when $a \geqslant \frac{6}{5}$, the original inequality has integer roots $(2,1,1)$, but $(x, y, z)$ cannot form a triangle, so $a<\frac{6}{5}$.

When $a<1$, $x^{2}+y^{2}+z^{2} \leqslant a(x y+y z+z x)<x y+y z+z x$, which is a contradiction! Hence, $a \geqslant 1$. We now prove that when $1 \leqslant a<\frac{6}{5}$, all positive integer roots of the original inequality can form the sides of a triangle.

First, $(1,1,1)$ is a solution to the original inequality, and they can form a triangle. If the inequality has integer roots $\left(x_{1}, y_{1}, z_{1}\right)$, and they cannot form the sides of a triangle, then, transforming the original equation: (assuming $\left.x_{1} \geqslant y_{1}+z_{1}\right) x^{2}-a(y+z) x+y^{2}+z^{2}-a y z \leqslant 0$.

Consider the function $f(x)=x^{2}-a\left(y_{1}+z_{1}\right) x+y_{1}^{2}+z_{1}^{2}-a y_{1} z_{1}$, its axis of symmetry is $x=$ $\frac{a}{2}\left(y_{1}+z_{1}\right)<\frac{3}{5}\left(y_{1}+z_{1}\right)<y_{1}+z_{1} \leqslant x_{1}$, so $\left(y_{1}+z_{1}, y_{1}, z_{1}\right)$ is also a solution to the original inequality.

Therefore, $a \geqslant \frac{\left(y_{1}+z_{1}\right)^{2}+y_{1}^{2}+z_{1}^{2}}{\left(y_{1}+z_{1}\right) y_{1}+\left(y_{1}+z_{1}\right) z_{1}+y_{1} z_{1}}=u$, then $(u-2) y_{1}^{2}+(3 u-$ 2) $y_{1} z_{1}+(u-2) z_{1}^{2}=0$

Clearly, $u \neq 2$, so $\Delta=(3 u-2)^{2}-4(u-2)^{2} \geqslant 0$, thus $u \geqslant \frac{6}{5}$, then $a \geqslant u \geqslant$ $\frac{6}{5}$, which is a contradiction!