Є. Let $x_{1}, x_{2}, y_{1}, y_{2} \in \mathbf{R}^{+}, a=\sqrt{x_{1} x_{2}}+$ $\sqrt{y_{1} y_{2}}, b=\sqrt{\left(x_{1}+y_{1}\right)\left(x_{2}+y_{2}\right)}$. Then the relationship between $a, b$ is $\qquad$ .
$\begin{array}{l}6 . a \leqslant b . \\ \begin{aligned} \frac{a}{b}= & \sqrt{\frac{x_{1} x_{2}}{\left(x_{1}+y_{1}\right)\left(x_{2}+y_{2}\right)}} \\ & +\sqrt{\frac{y_{1} y_{2}}{\left(x_{1}+y_{1}\right)\left(x_{2}+y_{2}\right)}} \\ \leqslant & \frac{1}{2}\left(\frac{x_{1}}{x_{1}+y_{1}}+\frac{x_{2}}{x_{2}+y_{2}}\right) \\ & +\frac{1}{2}\left(\frac{y_{1}}{x_{1}+y_{1}}+\frac{y_{2}}{x_{2}+y_{2}}\right) \\ = & 1 .\end{aligned} \\ \therefore a \leqslant b .\end{array}$