Problem 1600

Prove by introducing parameter $\lambda$, construct the inequality
$$\frac{1}{1+(n-1) a_{1}} \geqslant \frac{a_{1}^{\lambda}}{a_{1}^{\lambda}+a_{2}^{\lambda}+\cdots+a_{n}^{\lambda}}$$

Below, we will study the feasibility of inequality (2).
Given $a_{1}>0, n \geqslant 3$, after rearrangement, inequality (2) is equivalent to
$$a_{2}^{\lambda}+a_{3}^{\lambda}+\cdots+a_{n}^{\lambda} \geqslant(n-1) a_{1}^{\lambda}+1$$

By the AM-GM inequality, we get $a_{2}^{\lambda}+a_{3}^{\lambda}+\cdots+a_{n}^{\lambda} \geqslant(n-1) \left(a_{2} a_{3} \cdots a_{n}\right)^{\frac{\lambda}{n-1}}$, and since $a_{1} a_{2} a_{3} \cdots a_{n}=1$, we have $(n-1) \cdot \left(a_{2} a_{3} \cdots a_{n}\right)^{\frac{\lambda}{n-1}}=(n-1) a_{1}^{-\frac{\lambda}{n-1}}$, thus we obtain
$$a_{2}^{\lambda}+a_{3}^{\lambda}+\cdots+a_{n}^{\lambda} \geqslant(n-1) a_{1}^{-\frac{\lambda}{n-1}}$$

Comparing (3) and (4), let $-\frac{\lambda}{n-1}=\lambda+1$, solving for $\lambda$ gives $\lambda=-\frac{n-1}{n}$, hence, there exists a parameter $\lambda=-\frac{n-1}{n}$, such that inequality (2) holds.

Similarly, when $i=1,2,3, \cdots, n$, the inequality