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 $\qquad$
$$
2 x^{4}+4 y^{4}+9=2 x^{4}+2+2+2+4 y^{4}+1+1+1 \geqslant 4 \sqrt[4]{16 x^{4}}+4 \sqrt[4]{4 y^{4}} \geqslant 8 x+4 \sqrt{2} y
$$

Therefore, $\frac{2 x+\sqrt{2} y}{2 x^{4}+4 y^{4}+9} \leqslant \frac{1}{4}$, with equality holding when $x=1, y=\frac{\sqrt{2}}{2}$.