Question 5 Find all positive integers $k$, such that for any positive numbers $a, b, c$ satisfying $abc=1$, the following inequality holds:
$$\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+3 k \geqslant(k+1)(a+b+c)$$
Let $a=b=\frac{1}{n+1}, c=(n+1)^{2}\left(n \in \mathbf{N}^{*}\right)$,
from (11) we get
$$k \leqslant \frac{n^{2}+2 n+1+\frac{1}{(n+1)^{4}}-\frac{2}{n+1}}{n^{2}+2 n+\frac{2}{n+1}-1}$$

and $\lim _{n+\infty} \frac{n^{2}+2 n+1+\frac{1}{(n+1)^{4}}-\frac{2}{n+1}}{n^{2}+2 n+\frac{2}{n+1}-1}=1$,
so $k \leqslant 1$, since $k$ is a positive integer, it can only be $k=1$, at this time the inequality (11) becomes
$$\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+3 \geqslant 2(a+b+c)$$

Let $x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}$, then $x, y, z$ also satisfy
$x y z=1$, the inequality (12) is equivalent to
$$\begin{array}{c}
x^{2}+y^{2}+z^{2}+3 \geqslant 2(x y+y z+z x) \\
\Leftrightarrow(x+y+z)\left(x^{2}+y^{2}+z^{2}+3\right) \geqslant 2(x y+y z \\
+z x)(x+y+z) \\
\quad \Leftrightarrow x^{3}+y^{3}+z^{3}+3(x+y+z) \geqslant x^{2} y+x y^{2}+ \\
y^{2} z+y z^{2}+z^{2} x+z x^{2}+6
\end{array}$$

By the variant (3) and $x+y+z \geqslant 3 \sqrt[3]{x y z}=3$., we know that (13) holds.