48. Find all positive real numbers $\lambda$ such that for any integer $n \geqslant 2$ and positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfying $\sum_{i=1}^{n} a_{i}=n$, we always have $\sum_{i=1}^{n} \frac{1}{a_{i}}-\lambda \prod_{i=1}^{n} \frac{1}{a_{i}} \leqslant n-\lambda$. (2010 China National Training Team Test
48. We need the following conclusion: the sequence $\left(1+\frac{1}{n}\right)^{n}$ is strictly monotonically increasing, and $\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n}=$ e:

We first prove that the inequality always holds when $\lambda \geqslant e$.
Assume $a_{n-1}=\min _{1 \leqslant i \leqslant n} a_{i}, a_{n}=\max _{1 \leqslant i \leqslant n} a_{\text {, }}$, then
$$\sum_{i=1}^{n} \frac{1}{a_{i}}-\lambda \prod_{i=1}^{n} \frac{1}{a_{i}}=\sum_{i=1}^{n-2} \frac{1}{a_{i}}+\frac{1}{a_{n} a_{n-1}}\left(n-\sum_{i=1}^{n-2} \frac{1}{a_{i}}-\frac{\lambda}{\prod_{i=1}^{n-2} \frac{1}{a_{i}}}\right)$$

By the arithmetic mean $\geqslant$ geometric mean, we get
$$\sum_{i=1}^{n-2} a_{i}=+\frac{\lambda}{\prod_{i=1}^{n-2}} a_{i} \geqslant(n-1)-\sqrt[n-1]{\lambda}$$

So
$$n-\sum_{i=1}^{n-2} a_{i}-\frac{\lambda}{\prod_{i=1}^{n-2} a_{i}} \leq n-(n-1) \sqrt[n-1]{\lambda}\left(\frac{n}{n-1}\right)^{n-1}=\left(1+\frac{1}{n-1}\right)^{n-1}$$, by the lemma, it holds when $\lambda \geqslant$ e > $\left(1+\frac{1}{n-1}\right)^{n+1}$.

Thus, keeping $a_{n}+a_{n-1}$ unchanged, making $a_{n} a_{n+1}$ larger (equivalent to bringing $a_{n-1}, a_{n}$ closer), expression (1) will not decrease. Since $a_{n} \leqslant 1 \leqslant a_{n}$ and $a_{n-1} a_{n} \leqslant 1 \cdot\left(a_{n}+a_{n-1}-1\right)$, when we adjust $a_{n-1}, a_{n}$ to 1 and $a_{n-1}+a_{n}-1$ respectively, expression (1) does not decrease. Hence
$$\sum_{i=1}^{n} \frac{1}{a_{i}}-\lambda \prod_{i=1}^{n} \frac{1}{a_{i}} \leqslant \sum_{i=1}^{n} 1-\lambda \prod_{i=1}^{n} 1=n-\lambda$$

On the other hand, for any $\lambda\lambda$. Take $a_{1}=$ $a_{2}=\cdots=a_{n-2}=\sqrt[n-2]{\lambda}$, then $a_{1}+a_{2}+\cdots+a_{n-2}=(n-2) \sqrt[n-1]{\lambda}$ and
$$\left(\frac{n}{n-2}\right)^{n-1}=\left(1+\frac{2}{n-2}\right)^{n-1}>\left(1+\frac{1}{n-2}\right)^{n-2}>\lambda$$

Thus
$$a_{1}+a_{2}+\cdots+a_{n-2}\left(1+\frac{1}{n-2}\right)^{n-2}>\lambda$$

At this point, keeping $a_{n-1}+a_{n}=n-(n-2) \sqrt[n-1]{\lambda}(>0)$ unchanged, $a_{n-1} \cdot a_{n}$ decreases, expression (1) will increase. Especially when $a_{n-1} \cdot a_{n} \rightarrow 0$, the right side of expression (1) tends to infinity. Therefore, when $\lambda<\mathrm{e}$, $\sum_{i=1}^{n} \frac{1}{a_{i}}-$ $\lambda \prod_{i=1}^{n} \frac{1}{a_{i}}$ has no upper bound.