设整数 $n\geqslant 3$，$\alpha,\beta,\gamma\in(0,1)$，$a_k , b_k , c_k \geqslant 0$（$k=1,2,\ldots,n$）满足
$\displaystyle
\sum\limits_{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
$
若对任意满足上述条件的 $a_k , b_k , c_k$（$k=1,2,\ldots,n$），均有 $\displaystyle \sum\limits_{k=1}^{n}(k+\lambda) a_{k} b_{k} c_{k} \leqslant \lambda$，求 $\lambda$ 的最小值．
$\displaystyle
\sum\limits_{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
$
若对任意满足上述条件的 $a_k , b_k , c_k$（$k=1,2,\ldots,n$），均有 $\displaystyle \sum\limits_{k=1}^{n}(k+\lambda) a_{k} b_{k} c_{k} \leqslant \lambda$，求 $\lambda$ 的最小值．
【难度】
【出处】
2013中国东南数学奥林匹克试题
【标注】
【答案】
略
【解析】
令 $a_{1}=\dfrac{\alpha}{1+\alpha}, b_{1}=\dfrac{\beta}{1+\beta}, c_{1}=\dfrac{\gamma}{1+\gamma}$，$a_i , b_i , c_i = 0$（$i=2,3,\ldots, n$），此时条件成立，故 $\lambda$ 须满足
$
(1+\lambda) \dfrac{\alpha}{1+\alpha} \cdot \dfrac{\beta}{1+\beta} \cdot \dfrac{\gamma}{1+\gamma} \leqslant \lambda
$
解得
$
\lambda \geqslant \dfrac{a \beta \gamma}{(1+\alpha)(1+\beta)(1+\gamma)-\alpha \beta \gamma}
$
记 $\dfrac{a \beta \gamma}{(1+\alpha)(1+\beta)(1+\gamma)-\alpha \beta \gamma}=\lambda_0$ 下面证明，对任意满足条件的 $a_k , b_k , c_k$，$k=1,2,\ldots,n$，有
$\displaystyle \sum\limits_{k=1}^{n}\left(k+\lambda_{0}\right) a_{k} b_{k} c_{k} \leqslant \lambda_{0}
$ ①
由题目条件知
$\begin{aligned} & \sum_{k=1}^{n}\left(\frac{k+\alpha}{\alpha} a_{k} \cdot \frac{k+\beta_{b}}{\beta} \cdot \frac{k+\gamma}{\gamma} c_{k}\right)^{\frac{1}{3}} \\ \leqslant & \left(\sum_{k=1}^{n} \frac{k+\alpha}{\alpha} a_{k}\right)^{\frac{1}{3}} \cdot\left(\sum_{k=1}^{n} \frac{k+\beta_{k}}{\beta} b_{k}\right)^{\frac{1}{3}} \cdot\left(\sum_{k=1}^{n} \frac{k+\gamma}{\gamma} c_{k}\right)^{\frac{1}{3}} \leqslant 1 \end{aligned}$
这里用到了结论：当 $x_i , y_i , z_i \geqslant 0$（$i=1,2,\ldots,n$）时，有
$\displaystyle \left(\sum\limits_{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)
$ ②
（为完整起见，我们将 ② 的证明过程附在最后．）
因此，为证 ①，只需证明对 $k=1,2,\ldots,n$，有
$
\dfrac{k+\lambda_{0}}{\lambda_{0}} a_{k} b_{k} c_{k} \leqslant\left(\dfrac{k+\alpha}{\alpha} \cdot \dfrac{k+\beta}{\beta} \cdot \dfrac{k+\gamma}{\gamma} \cdot a_{k} b_{k} c_{k}\right)^{\frac{1}{3}}
$
即 $\dfrac{k+\lambda_{0}}{\lambda_{0}}\left(a_{k} b_{k} c_{k}\right)^{\frac{2}{3}} \leqslant\left(\dfrac{(k+\alpha)(k+\beta)(k+\gamma)}{a \beta \gamma}\right)^{\frac{1}{3}}
$ ③
事实上，
$\begin{aligned} \lambda_{0} &=\frac{a \beta \gamma}{1+(\alpha+\beta+\gamma)+\left(\alpha \beta+\beta \gamma+\gamma_{\alpha}\right)} \\ & \geqslant \frac{a \beta \gamma}{k^{2}+(\alpha+\beta+\gamma) k+\left(\alpha \beta+\beta \gamma+\gamma_{\alpha}\right)} \\ &=\frac{k \alpha \beta \gamma}{(k+\alpha)(k+\beta)(k+\gamma)-\alpha \beta \gamma} \end{aligned}$
因此 $
\dfrac{k+\lambda_{0}}{\lambda_{0}} \leqslant \dfrac{(k+\alpha)(k+\beta)(k+\gamma)}{\alpha \beta \gamma}
$ ④
又由于 $(k+\alpha) a_{k} \leqslant \alpha,(k+\beta) b_{k} \leqslant \beta,(k+\gamma) c_{k} \leqslant \gamma$，故
$
\left(a_{k} b_{k} c_{k}\right)^{\dfrac{2}{3}} \leqslant\left(\dfrac{\alpha \beta \gamma}{(k+\alpha)(k+\beta)(k+\gamma)}\right)^{\frac{2}{3}}
$ ⑤
由 ④，⑤ 可知 ③ 成立，从而 ① 成立．
综上所述，$\lambda_{\min }=\lambda_{0}=\dfrac{a \beta \gamma}{(1+\alpha)(1+\beta)(1+\gamma)-\alpha \beta \gamma}$．
$
(1+\lambda) \dfrac{\alpha}{1+\alpha} \cdot \dfrac{\beta}{1+\beta} \cdot \dfrac{\gamma}{1+\gamma} \leqslant \lambda
$
解得
$
\lambda \geqslant \dfrac{a \beta \gamma}{(1+\alpha)(1+\beta)(1+\gamma)-\alpha \beta \gamma}
$
记 $\dfrac{a \beta \gamma}{(1+\alpha)(1+\beta)(1+\gamma)-\alpha \beta \gamma}=\lambda_0$ 下面证明，对任意满足条件的 $a_k , b_k , c_k$，$k=1,2,\ldots,n$，有
$\displaystyle \sum\limits_{k=1}^{n}\left(k+\lambda_{0}\right) a_{k} b_{k} c_{k} \leqslant \lambda_{0}
$ ①
由题目条件知
$\begin{aligned} & \sum_{k=1}^{n}\left(\frac{k+\alpha}{\alpha} a_{k} \cdot \frac{k+\beta_{b}}{\beta} \cdot \frac{k+\gamma}{\gamma} c_{k}\right)^{\frac{1}{3}} \\ \leqslant & \left(\sum_{k=1}^{n} \frac{k+\alpha}{\alpha} a_{k}\right)^{\frac{1}{3}} \cdot\left(\sum_{k=1}^{n} \frac{k+\beta_{k}}{\beta} b_{k}\right)^{\frac{1}{3}} \cdot\left(\sum_{k=1}^{n} \frac{k+\gamma}{\gamma} c_{k}\right)^{\frac{1}{3}} \leqslant 1 \end{aligned}$
这里用到了结论：当 $x_i , y_i , z_i \geqslant 0$（$i=1,2,\ldots,n$）时，有
$\displaystyle \left(\sum\limits_{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)
$ ②
（为完整起见，我们将 ② 的证明过程附在最后．）
因此，为证 ①，只需证明对 $k=1,2,\ldots,n$，有
$
\dfrac{k+\lambda_{0}}{\lambda_{0}} a_{k} b_{k} c_{k} \leqslant\left(\dfrac{k+\alpha}{\alpha} \cdot \dfrac{k+\beta}{\beta} \cdot \dfrac{k+\gamma}{\gamma} \cdot a_{k} b_{k} c_{k}\right)^{\frac{1}{3}}
$
即 $\dfrac{k+\lambda_{0}}{\lambda_{0}}\left(a_{k} b_{k} c_{k}\right)^{\frac{2}{3}} \leqslant\left(\dfrac{(k+\alpha)(k+\beta)(k+\gamma)}{a \beta \gamma}\right)^{\frac{1}{3}}
$ ③
事实上，
$\begin{aligned} \lambda_{0} &=\frac{a \beta \gamma}{1+(\alpha+\beta+\gamma)+\left(\alpha \beta+\beta \gamma+\gamma_{\alpha}\right)} \\ & \geqslant \frac{a \beta \gamma}{k^{2}+(\alpha+\beta+\gamma) k+\left(\alpha \beta+\beta \gamma+\gamma_{\alpha}\right)} \\ &=\frac{k \alpha \beta \gamma}{(k+\alpha)(k+\beta)(k+\gamma)-\alpha \beta \gamma} \end{aligned}$
因此 $
\dfrac{k+\lambda_{0}}{\lambda_{0}} \leqslant \dfrac{(k+\alpha)(k+\beta)(k+\gamma)}{\alpha \beta \gamma}
$ ④
又由于 $(k+\alpha) a_{k} \leqslant \alpha,(k+\beta) b_{k} \leqslant \beta,(k+\gamma) c_{k} \leqslant \gamma$，故
$
\left(a_{k} b_{k} c_{k}\right)^{\dfrac{2}{3}} \leqslant\left(\dfrac{\alpha \beta \gamma}{(k+\alpha)(k+\beta)(k+\gamma)}\right)^{\frac{2}{3}}
$ ⑤
由 ④，⑤ 可知 ③ 成立，从而 ① 成立．
综上所述，$\lambda_{\min }=\lambda_{0}=\dfrac{a \beta \gamma}{(1+\alpha)(1+\beta)(1+\gamma)-\alpha \beta \gamma}$．
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