Problem 1484

74. Let $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}$, the original inequality is equivalent to
$$(x+y)\left(\frac{a_{1}}{a_{2} x+a_{3} y}+\frac{a_{2}}{a_{3} x+a_{4} y}+\cdots+\frac{a_{n-1}}{a_{n} x+a_{1} y}+\frac{a_{n}}{a_{1} x+a_{2} y} \geqslant n\right.$$

Let $a_{n+1}=a_{1}, a_{n+2}=a_{2}$, by the AM-GM inequality we have
$$\sum_{i=1}^{n} \frac{a_{i} x+a_{i+1} y}{a_{i+1} x+a_{i+2} y} \geqslant n \sqrt[n]{\prod_{i=1}^{n} \frac{a_{i} x+a_{i+1} y}{a_{i+1} x+a_{i+2} y}}=n$$

It suffices to prove
$$(x+y) \sum_{i=1}^{n} \frac{a_{i}}{a_{i+1} x+a_{i+2} y} \geqslant \sum_{i=1}^{n} \frac{a_{i} x+a_{i+1} y}{a_{i+1} x+a_{i+2} y}$$

And
$$\begin{array}{l} (x+y) \sum_{i=1}^{n} \frac{a_{i}}{a_{i+1} x+a_{i+2} y}-\sum_{i=1}^{n} \frac{a_{i} x+a_{i+1} y}{a_{i+1} x+a_{i+2} y}= \\ \sum_{i=1}^{n} \frac{\left(a_{i}-a_{i+1}\right) y}{a_{i+1} x+a_{i+2} y}= \\ y \sum_{i=1}^{n} \frac{a_{i}-a_{i+1}}{a_{i+1} x+a_{i+2} y} \end{array}$$

So it suffices to prove
$$\sum_{i=1}^{n} \frac{a_{i}-a_{i+1}}{a_{i+1} x+a_{i+2} y} \geqslant 0$$

Also
$$\begin{aligned} \sum_{i=1}^{n} \frac{a_{i}-a_{i+1}}{a_{i+1} x+a_{i+2} y}= & \sum_{i=1}^{n-1} \frac{a_{i}-a_{i+1}}{a_{i+1} x+a_{i+2} y}-\frac{a_{1}-a_{n}}{a_{1} x+a_{2} y}= \\ & \sum_{i=1}^{n-1} \frac{a_{i}-a_{i+1}}{a_{i+1} x+a_{i+2} y}-\sum_{i=1}^{n-1} \frac{a_{i}-a_{i+1}}{a_{1} x+a_{2} y}= \\ & \sum_{i=1}^{n-1}\left(a_{i}-a_{i+1}\right)\left(\frac{1}{a_{i+1} x+a_{i+2} y}-\frac{1}{a_{1} x+a_{2} y}\right) \end{aligned}$$

Since $a_{i} \geqslant a_{i+1}, i=1,2, \cdots, n-1$, for $i=1,2, \cdots, n-2$, we have $a_{i}-a_{i+1} \geqslant 0$, $a_{1} \geqslant a_{i+1}, a_{2} \geqslant a_{i+2}$, thus,
$$\frac{1}{a_{i+1} x+a_{i+2} y} \geqslant \frac{1}{a_{1} x+a_{2} y}$$

At this point $\left(a_{i}-a_{i+1}\right)\left(\frac{1}{a_{i+1} x+a_{i+2} y}-\frac{1}{a_{1} x+a_{2} y}\right) \geqslant 0, i=1,2, \cdots, n-2$.
For $i=n-1, a_{n-1}-a_{n} \geqslant 0$, given condition $\frac{x}{y} \geqslant \frac{a_{1}-a_{2}}{a_{1}-a_{n}} \Leftrightarrow\left(a_{1}-a_{n}\right) x \geqslant\left(a_{1}-\right.$ $\left.a_{2}\right) y \Leftrightarrow a_{1} x+a_{2} y \geqslant a_{n} x+a_{1} y \Leftrightarrow \frac{1}{a_{n} x+a_{1} y} \geqslant \frac{1}{a_{1} x+a_{2} y}$, so ( $a_{n-1}-$
$$\left.a_{n}\right)\left(\frac{1}{a_{n} x+a_{1} y}-\frac{1}{a_{1} x+a_{2} y}\right) \geqslant 0$$

In summary, $\sum_{i=1}^{n-1}\left(a_{i}-a_{i+1}\right)\left(\frac{1}{a_{i+1} x+a_{i+2} y}-\frac{1}{a_{1} x+a_{2} y}\right) \geqslant 0$. Therefore, the original inequality holds.