3. Let $n$ be a natural number, for any real numbers $x, y, z$ there is always $\left(x^{2}+y^{2}+z^{2}\right) \leqslant$ $n\left(x^{4}+y^{4}+z^{4}\right)$, then the minimum value of $n$ is $\qquad$
3. Let $a=x^{2}, b=y^{2}, c=z^{2}$, the given inequality becomes
$$
(a+b+c)^{2} \leqslant n\left(a^{2}+b^{2}+c^{2}\right) .
$$

On one hand,
$$
\begin{aligned}
& (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 a c \\
\leqslant & a^{2}+b^{2}+c^{2}+\left(a^{2}+b^{2}\right)+\left(b^{2}+c^{2}\right)+\left(a^{2}+c^{2}\right) \\
= & 3\left(a^{2}+b^{2}+c^{2}\right),
\end{aligned}
$$

so when $x=3$, the inequality holds.
On the other hand, when $a=b=c>0$, the given inequality becomes $9 a^{2} \leqslant 3 n a^{2}$, which implies $n \geqslant 3$. Therefore, the minimum value of $n$ is 3.