1. Positive real numbers $x, y$ satisfy $x y=1$. Then, $\frac{1}{x^{4}}+$ $\frac{1}{4 y^{4}}$ has the minimum value of ( ).
(A) $\frac{1}{2}$
(B) $\frac{5}{8}$
(C) 1
(D) $\sqrt{2}$
$-1 .(\mathrm{C})$.
Since $x=\frac{1}{y}$, we have,
$$
\frac{1}{x^{4}}+\frac{1}{4 y^{4}}=\frac{1}{x^{4}}+\frac{x^{4}}{4}=\left(\frac{1}{x^{2}}-\frac{x^{2}}{2}\right)^{2}+1 \text {. }
$$

When $\frac{1}{x^{2}}=\frac{x^{2}}{2}$, i.e., $x=\sqrt[4]{2}$, $\frac{1}{x^{4}}+\frac{1}{4 y^{4}}$ is minimized, and the minimum
value is 1.