3. Let $x$ and $y$ be positive numbers, and $x+y=4$. If the inequality $\frac{1}{x}+\frac{4}{y} \geqslant m$ always holds, then the maximum value of the real number $m$ is $\qquad$
$\begin{array}{l}\text { 3. } \frac{9}{4} \\ \frac{1}{x}+\frac{4}{y}=\frac{1}{4}(x+y)\left(\frac{1}{x}+\frac{4}{y}\right) \\ =\frac{1}{4}\left(5+\frac{y}{x}+\frac{4 x}{y}\right) \\ \geqslant \frac{1}{4}\left(5+2 \sqrt{\frac{y}{x} \cdot \frac{4 x}{y}}\right)=\frac{9}{4} .\end{array}$