1. Given $t \in \mathbf{R}_{+}$. Then the minimum value of $\frac{1}{\sqrt{1+3 t^{4}}}+\frac{t^{3}}{\sqrt[4]{12}}$ is $\qquad$ .
$$
=1 \cdot \frac{2 \sqrt{2}}{3} \text {. }
$$

From $1+3 t^{4} \geqslant 2 \sqrt{3 t^{4}}=\sqrt{12 t^{4}}$, we know
$$
\begin{array}{l}
\sqrt{1+3 t^{4}} \geqslant \sqrt{12 t} . \\
\text { Therefore, } \frac{1}{\sqrt{1+3 t^{4}}}+\frac{t^{3}}{\sqrt{12}} \geqslant \frac{1+t^{4}}{\sqrt{1+3 t^{4}}} . \\
=\frac{\sqrt{1+3 t^{4}}}{3}+\frac{2}{3 \sqrt{1+3 t^{4}}} \geqslant 2 \sqrt{\frac{2}{9}}=\frac{2 \sqrt{2}}{3},
\end{array}
$$

The equality holds if and only if $1+3 t^{4}=2$, i.e., $t=\frac{\sqrt{27}}{3}$.