Let $x_{1}, x_{2}$ be different positive real numbers, $a$ be a real number in $(0,1)$, $y_{1}=\frac{x_{1}}{1+a}+\frac{a x_{2}}{1+a}, y_{2}=\frac{x_{2}}{1+a}+$ $\frac{a x_{1}}{1+a}$, then the relationship between $x_{1} x_{2} \leftrightarrows y_{1} y_{2}$ is
(A) $x_{1} \cdot x_{2}>y_{1} y_{2}$
(B) $x_{1} x_{2}=y_{1} y_{2}$
(C) $x_{1} x_{2}<y_{1} y_{2}$
(D) Cannot be determined, depends on the value of $a$
2. (C).

Note that, $y_{1} y_{2}=\frac{1+a^{2}}{(1+a)^{2}} x_{1} x_{2}+\frac{1}{(1 \div a)^{2}}\left(\left(a x_{1}\right)^{2}+\left(a x_{2}\right)^{2}\right)>\frac{1+a^{2}}{(1+a)^{2}} x_{1} x_{2}+$ $\frac{2 a}{(1+a)^{2}} x_{1} x_{2}=x_{1} x_{2}$ (here the strict inequality is due to $a x_{1}=u x_{2}$ leading to $x_{1}=x_{2}$, which contradicts the assumption that $x_{1}, x_{2}$ are distinct).