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. 【Analysis and Solution】Solution 1 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, $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 a c$
$\leqslant a^{2}+b^{2}+c^{2}+\left(a^{2}+b^{2}\right)+\left(b^{2}+c^{2}\right)+\left(a^{2}+c^{2}\right)$
$=3\left(a^{2}+b^{2}+c^{2}\right)$, so when $x=3$, the inequality holds;
On the other hand, when $a=b=c>0$, the given inequality simplifies to $9 a^{2} \leqslant 3 n a^{2}$, which implies $n \geqslant 3$, hence the minimum value of $n$ is 3.
Solution 2 $3\left(x^{4}+y^{4}+z^{4}\right)-\left(x^{2}+y^{2}+z^{2}\right)^{2}$
$=\left(x^{2}-y^{2}\right)^{2}+\left(y^{2}-z^{2}\right)+\left(z^{2}-x^{2}\right)^{2} \geqslant 0$. Or, directly by the Cauchy-Schwarz inequality,
$3\left(x^{4}+y^{4}+z^{4}\right)=(1+1+1)\left(x^{4}+y^{4}+z^{4}\right) \geqslant\left(x^{2}+y^{2}+z^{2}\right)^{2}$.