1. For $x_{1}>x_{2}>0,1>a>0$, let
$$
y_{1}=\frac{x_{1}}{1+a}+\frac{a x_{2}}{1+a}, y_{2}=\frac{a x_{1}}{1+a}+\frac{x_{2}}{1+a} \text {. }
$$

Then the relationship between $x_{1} x_{2}$ and $y_{1} y_{2}$ is ( ).
(A) $x_{1} x_{2}>y_{1} y_{2}$
(B) $x_{1} x_{2}=y_{1} y_{2}$
(C) $x_{1} x_{2}<y_{1} y_{2}$
(D) Cannot be determined, depends on $a$
-1. (C).
Path 1: (Arithmetic Mean Inequality)
$$
\begin{aligned}
y_{1} y_{2} & =\frac{a}{(1+a)^{2}}\left(x_{1}^{2}+x_{2}^{2}\right)+\frac{1+a^{2}}{(1+a)^{2}} x_{1} x_{2} \\
& >\frac{a}{(1+a)^{2}}\left(2 x_{1} x_{2}\right)+\frac{1+a^{2}}{(1+a)^{2}} x_{1} x_{2} \\
& =x_{1} x_{2} .
\end{aligned}
$$

Path 2: (Vicsek Inequality)
$$
\begin{array}{l}
y_{1} y_{2}=\left[\left(\sqrt{\frac{x_{1}}{1+a}}\right)^{2}+\left(\sqrt{\frac{a x_{2}}{1+a}}\right)^{2}\right] \\
- \left[\left(\sqrt{\frac{x_{2}}{1+a}}\right)^{2}+\left(\sqrt{\frac{a x_{1}}{1+a}}\right)^{2}\right] \\
>\left(\frac{1}{1+a}-\sqrt{x_{1} x_{2}}+\frac{a}{1+x_{2}}-\sqrt{x_{2} x_{2}}\right)^{2}=x_{1} x_{2} . \\
\end{array}
$$