《2025年全国重点高中提前招生同步强化全真试卷八年级数学上册》

20. (华中师大一中附预录)已知整数$a,b,c,d$满足$abcd = 6(a - 1)(b - 1)(c - 1)(d - 1)$.
(1)是否存在满足上述条件的$a,b,c,d$均为整数?若存在,求出所有的解;若不存在,请说明理由;
(2)若$a\gt1,b\gt1,c\gt1,d\gt1$,求出$a + b + c + d$的最小值.
参考知识:①当$a,b,c,d$均大于$0$时,$abcd\leqslant(\frac{a + b + c + d}{4})^{4}$,等号当且仅当$a = b = c = d$时成立.②当$a,b,c,d$均大于$0$时,$(a + b + c + d)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})\geqslant16$,等号当且仅当$a = b = c = d$时成立.

答案： 20.解:
(1) 不妨设满足条件的整数$a \geq b \geq c \geq d$, 显然$d \geq 2$. 若$d \geq 3$, 这时$\frac{1}{a} \leq \frac{1}{b} \leq \frac{1}{c} \leq \frac{1}{d} \leq \frac{1}{3}$, 由$abcd = 6(a - 1)(b - 1)(c - 1)(d - 1)$, 可得$\frac{1}{6} = (1 - \frac{1}{a})(1 - \frac{1}{b})(1 - \frac{1}{c})(1 - \frac{1}{d}) \geq (\frac{2}{3})^4 = \frac{16}{81} ＞ \frac{1}{6}$, 矛盾.
∴ $d = 2$代入式中有$abc = 3(a - 1)(b - 1)(c - 1)$, 由上面易知$c \geq 2$. 若$c \geq 4$, 则$\frac{1}{a} \leq \frac{1}{b} \leq \frac{1}{c} \leq \frac{1}{4}$, 故$(1 - \frac{1}{a})(1 - \frac{1}{b})(1 - \frac{1}{c}) \geq (1 - \frac{1}{4})^3 ＞ \frac{3}{4}$. 当$c = 2$时, $2ab = 3(a - 1)(b - 1)$, 即$(a - 3)(b - 3) = 6$, 又$a \geq b \geq 3$, 故$\begin{cases} a - 3 = 6 \\ b - 3 = 1 \end{cases}$或$\begin{cases} a - 3 = 3 \\ b - 3 = 2 \end{cases}$, 解得$\begin{cases} a = 9 \\ b = 4 \end{cases}$或$\begin{cases} a = 6 \\ b = 5 \end{cases}$, 当$c = 3$时, $3ab = 6(a - 1)(b - 1)$, 即$(a - 2)(b - 2) = 2$, 又$a \geq b \geq 3$, 故$\begin{cases} a - 2 = 2 \\ b - 2 = 1 \end{cases}$, 解得$\begin{cases} a = 4 \\ b = 3 \end{cases}$, 故存在$a$, $b$, $c$, $d$的值为$(4, 3, 3, 2)$, $(9, 4, 2, 2)$, $(6, 5, 2, 2)$.
(2) 由$abcd = 6(a - 1)(b - 1)(c - 1)(d - 1)$, 可得$\frac{1}{6} = (1 - \frac{1}{a})(1 - \frac{1}{b})(1 - \frac{1}{c})(1 - \frac{1}{d}) \leq 4 - \frac{4}{\sqrt[4]{6}}$, 又$(a + b + c + d)(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}) \geq 16$, 则有$a + b + c + d \geq \frac{16}{4 - \sqrt[4]{6}} = \frac{4\sqrt[4]{6}}{\sqrt[4]{6} - 1}$, 故$a + b + c + d$的最小值为$\frac{4\sqrt[4]{6}}{\sqrt[4]{6} - 1}$, 当且仅当$a = b = c = d$时, 取得此最小值.