9.4 Let $m=\left(\frac{1}{3}\right)^{\frac{1}{5}}, n=\left(\frac{1}{4}\right)^{\frac{1}{3}}, \quad p=\left(\frac{1}{5}\right)^{\frac{1}{4}}$. Then their size relationship is
(A) $m<n<p$.
(B) $m<p<n$.
(C) $n<p<m$.
(D) $p<n<m$.
(China Partial Provinces and Cities Junior High School Mathematics Correspondence Competition, 1989)
[Solution 1] Given $m=\left(\frac{1}{3}\right)^{\frac{1}{5}}=\left(\frac{1}{3^{4}}\right)^{\frac{1}{20}}$, and
$$
p=\left(\frac{1}{5}\right)^{\frac{1}{4}}=\left(\frac{1}{5^{5}}\right)^{\frac{1}{20}},
$$

since $3^{4}5^{3}$, thus $n<p$.
Therefore, $n<p<m$.
Hence, the answer is $(C)$.
[Solution 2] Let $\lg m=\frac{1}{5}(\lg 1-\lg 3)=-\frac{1}{5} \lg 3$, and
$$
\lg n=-\frac{2}{3} \lg 2, \quad \lg p=\frac{1}{4}(\lg 2-1).
$$

Given $\lg 2=0.3010 \cdots, \lg 3=0.4771 \cdots$.
We know $\lg n<\lg p<\lg m$, thus $n<p<m$.
Hence, the answer is $(C)$.