$13 \cdot 39$ For all real numbers $x, y$, and $z$, then
$$
\left(x^{2}+y^{2}+z^{2}\right)^{2} \leqslant n\left(x^{4}+y^{4}+z^{4}\right) \text {. }
$$

the smallest integer $n$ is
(A) 2 .
(B) 3 .
(C) 4 .
(D) 6 .
(E) There is no such integer $n$.
(28th American High School Mathematics Examination, 1977)
[Solution] Let $a=x^{2}, \quad b=y^{2}, \quad c=z^{2}$, then

we have
$$
0 \leqslant(a-b)^{2}+(b-c)^{2}+(c-a)^{2},
$$

or
$$
\begin{array}{c}
a b+b c+c a \leqslant a^{2}+b^{2}+c^{2}, \\
\frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}} \leqslant 1,
\end{array}
$$

Multiplying both sides of the inequality by 2 and adding 1, we get

that is $\square$
$$
\frac{a^{2}+b^{2}+c^{2}+2(a b+b c+c a)}{a^{2}+b^{2}+c^{2}} \leqslant 3,
$$

Therefore,
$$
(a+b+c)^{2} \leqslant 3\left(a^{2}+b^{2}+c^{2}\right) \text {. }
$$

Choosing $a=b=c>0$, from the assumption $(3 a)^{2} \leqslant n \cdot 3 a^{2}$, we get $n \geqslant 3$. Hence, the answer is $(B)$.