1. Given the inequality
$$
3 x+4 \sqrt{x y} \leqslant a(x+y)
$$

holds for all positive numbers $x$ and $y$. Then the minimum value of the real number $a$ is . $\qquad$
-1.4 .
From the problem, we know that $a \geqslant\left(\frac{3 x+4 \sqrt{x y}}{x+y}\right)_{\text {max }}$.
$$
\text { Also, } \frac{3 x+4 \sqrt{x y}}{x+y} \leqslant \frac{3 x+(x+4 y)}{x+y}=4 \text {, }
$$

with equality holding if and only if $x=4 y>0$.
Therefore, the minimum value of $a$ is 4.