8. Let integer $n \geqslant 3, \alpha, \beta, \gamma \in (0,1), a_{k}, b_{k}, c_{k} \geqslant 0 (k=1,2, \cdots, n)$ satisfy
$$
\begin{array}{l}
\sum_{k=1}^{n}(k+\alpha) a_{k} \leqslant \alpha, \\
\sum_{k=1}^{n}(k+\beta) b_{k} \leqslant \beta, \\
\sum_{k=1}^{n}(k+\gamma) c_{k} \leqslant \gamma .
\end{array}
$$

If for any $a_{k}, b_{k}, c_{k} (k=1,2, \cdots, n)$ satisfying the above conditions, we have
$$
\sum_{k=1}^{n}(k+\lambda) a_{k} b_{k} c_{k} \leqslant \lambda .
$$

Find the minimum value of $\lambda$.
(Qi Shenghong)
8. Let $a_{1}=\frac{\alpha}{1+\alpha}, b_{1}=\frac{\beta}{1+\beta}, c_{1}=\frac{\gamma}{1+\gamma}$, where $a_{i}, b_{i}, c_{i}=0(i=2,3, \cdots, n)$, at this point, the condition holds. Therefore, $\lambda$ must satisfy
$$
\begin{array}{l}
(1+\lambda) \frac{\alpha}{1+\alpha} \cdot \frac{\beta}{1+\beta} \cdot \frac{\gamma}{1+\gamma} \leqslant \lambda \\
\Rightarrow \lambda \geqslant \frac{\alpha \beta \gamma}{(1+\alpha)(1+\beta)(1+\gamma)-\alpha \beta \gamma} . \\
\text { Let } \frac{\alpha \beta \gamma}{(1+\alpha)(1+\beta)(1+\gamma)-\alpha \beta \gamma}=\lambda_{0} .
\end{array}
$$

Next, we prove: For any $a_{k}, b_{k}, c_{k}(k=1,2, \cdots, n)$ that satisfy the condition, we have
$$
\sum_{k=1}^{n}\left(k+\lambda_{0}\right) a_{k} b_{k} c_{k} \leqslant \lambda_{0} .
$$

From the given, we have
$$
\begin{array}{l}
\sum_{k=1}^{n}\left(\frac{k+\alpha}{\alpha} a_{k} \cdot \frac{k+\beta}{\beta} b_{k} \cdot \frac{k+\gamma}{\gamma} c_{k}\right)^{\frac{1}{3}} \\
\leqslant\left(\sum_{k=1}^{n} \frac{k+\alpha}{\alpha} a_{k}\right)^{\frac{1}{3}}\left(\sum_{k=1}^{n} \frac{k+\beta}{\beta} b_{k}\right)^{\frac{1}{3}}\left(\sum_{k=1}^{n} \frac{k+\gamma}{\gamma} c_{k}\right)^{\frac{1}{3}} \\
\leqslant 1,
\end{array}
$$

This is because when $x_{i}, y_{i}, z_{i} \geqslant 0(i=1,2, \cdots, n)$, we have
$$
\left(\sum_{i=1}^{n} x_{i} y_{i} z_{i}\right)^{3} \leqslant\left(\sum_{i=1}^{n} x_{i}^{3}\right)\left(\sum_{i=1}^{n} y_{i}^{3}\right)\left(\sum_{i=1}^{n} z_{i}^{3}\right) \text {. }
$$

Therefore, to prove equation (1), it suffices to prove that for $k=1,2, \cdots, n$, we have
$$
\frac{k+\lambda_{0}}{\lambda_{0}} a_{k} b_{k} c_{k} \leqslant\left(\frac{k+\alpha}{\alpha} \cdot \frac{k+\beta}{\beta} \cdot \frac{k+\gamma}{\gamma} a_{k} b_{k} c_{k}\right)^{\frac{1}{3}},
$$

i.e.,
$$
\frac{k+\lambda_{0}}{\lambda_{0}}\left(a_{k} b_{k} c_{k}\right)^{\frac{2}{3}} \leqslant\left[\frac{(k+\alpha)(k+\beta)(k+\gamma)}{\alpha \beta \gamma}\right]^{\frac{1}{3}} \text {. }
$$

In fact,
$$
\begin{array}{l}
\lambda_{0}=\frac{\alpha \beta \gamma}{1+(\alpha+\beta+\gamma)+(\alpha \beta+\beta \gamma+\gamma \alpha)} \\
\geqslant \frac{\alpha \beta \gamma}{k^{2}+(\alpha+\beta+\gamma) k+(\alpha \beta+\beta \gamma+\gamma \alpha)} \\
=\frac{k \alpha \beta \gamma}{(k+\alpha)(k+\beta)(k+\gamma)-\alpha \beta \gamma} . \\
\text { Hence } \frac{k+\lambda_{0}}{\lambda_{0}} \leqslant \frac{(k+\alpha)(k+\beta)(k+\gamma)}{\alpha \beta \gamma} . \\
\text { Also, }(k+\alpha) a_{k} \leqslant \alpha, \\
(k+\beta) b_{k} \leqslant \beta, \\
(k+\gamma) c_{k} \leqslant \gamma,
\end{array}
$$

then $\left(a_{k} b_{k} c_{k}\right)^{\frac{2}{3}}$
$$
\leqslant\left[\frac{\alpha \beta \gamma}{(k+\alpha)(k+\beta)(k+\gamma)}\right]^{\frac{2}{3}} .
$$

From equations (4) and (5), we know that equation (3) holds. Therefore, equation (1) holds.
In summary, $\lambda_{\min }=\lambda_{0}=\frac{\alpha \beta \gamma}{(1+\alpha)(1+\beta)(1+\gamma)-\alpha \beta \gamma}$.