Example 3 If $\left(\log _{2} 3\right)^{x}-\left(\log _{5} 3\right)^{r} \geqslant\left(\log _{2} 3\right)^{-v}$ $-\left(\log _{5} 3\right)^{-\nu}$, then $(\quad$.
(A) $x-y \geqslant 0$
(B) $x+y \geqslant 0$
(C) $x-y \leqslant 0$
(D) $x+y \leqslant 0$
Sol: $\because 0<\log _{5} 3<1<\log _{2} 3$,
$\therefore y_{1}=\left(\log _{2} 3\right)^{2}$ is an increasing function.
Also, $\because\left(\log _{5} 3\right)^{x}$ is a decreasing function,
$\therefore y_{2}=-\left(\log _{5} 3\right)^{\prime}$ is an increasing function.
Therefore, $y=y_{1}+y_{2}=\left(\log _{2} 3\right)^{\prime}-\left(\log _{5} 3\right)^{\prime}$ is
an increasing function.
Given $\left(\log _{2} 3\right)^{x}-\left(\log _{5} 3\right)^{x} \geqslant\left(\log _{2} 3\right)^{-v}$
$-\left(\log _{5} 3\right)^{-y}$, we get $x \geqslant-y$, i.e., $x+y \geqslant 0$. Therefore, the correct choice is
(B).