5. Let $x, y$ be real numbers. Then the maximum value of $\frac{2 x+\sqrt{2} y}{2 x^{4}+4 y^{4}+9}$ is . $\qquad$
5. $\frac{1}{4}$.
$$
\begin{array}{l}
\text { Given } 2 x^{4}+2+2+2 \geqslant 8 x, \\
4 y^{4}+1+1+1 \geqslant 4 \sqrt{2} y, \\
2 x^{4}+4 y^{4}+9 \geqslant 8 x+4 \sqrt{2} y \\
\Rightarrow \frac{2 x+\sqrt{2} y}{2 x^{4}+4 y^{4}+9} \leqslant \frac{1}{4} .
\end{array}
$$

When $x=1, y=\frac{\sqrt{2}}{2}$, the maximum value $\frac{1}{4}$ is achieved.