主元配方不等式

(2024年中国东南地区数学奥林匹克竞赛第1题)若实数\(\small \tau\)满足：对任意正整数\(\small x,y,z\)，均有

\[\small \small x^2+2y^2+4z^2+8\ge 2x(y+z+\tau), \]

则称\(\small \tau\)为“平生”数.记最大的平生数为\(\small \tau_0\).

(1)求\(\small \tau_0\)的值;

(2)求方程\(\small x^2+2y^2+4z^2+8= 2x(y+z+\tau_0)\)的所有正整数解\(\small (x,y,z)\).

解：(1)原不等式可等价转换为

\[\small \frac{x}{2}+\frac{y^2}{x}+\frac{2z^2}{x}+\frac{4}{x}-y-z\ge \tau, \]

注意到

\[\small \text{LHS}=\frac{1}{x}\left(y-\frac{x}{2}\right)^2+\frac{2}{x}\left(z-\frac{x}{4}\right)^2+\frac{x}{8}+\frac{4}{x}. \]

因为\(\small x,y,z\)是正整数，所以需要对\(x\)进行分类讨论：

(i)如果\(\small x\equiv 0\pmod 4\)，那么

\[\small \begin{aligned} \frac{1}{x}\left(y-\frac{x}{2}\right)^2+\frac{2}{x}\left(z-\frac{x}{4}\right)^2+\frac{x}{8}+\frac{4}{x}\\ \ge \frac{x}{8}+\frac{4}{x}\ge \frac{3}{2}, \end{aligned} \]

当且仅当\(\small (x,y,z)=(4,2,1)\)或\(\small (8,4,2)\)时等号成立;

(ii)如果\(\small x\equiv 1\pmod 4\)，那么

\[\small \begin{aligned} \frac{1}{x}\left(y-\frac{x}{2}\right)^2+\frac{2}{x}\left(z-\frac{x}{4}\right)^2+\frac{x}{8}+\frac{4}{x}\\ \ge \frac{1}{4x}+\frac{1}{8x}+\frac{x}{8}+\frac{4}{x}=\frac{35}{8x}+\frac{x}{8} \\ \ge \frac{3}{2}, \end{aligned} \]

当且仅当\(\small (x,y,z)=(5,2,1)\)或\(\small (5,3,1)\)时等号成立;

(iii)如果\(\small x\equiv 2\pmod 4\)，那么

\[\small \begin{aligned} \frac{1}{x}\left(y-\frac{x}{2}\right)^2+\frac{2}{x}\left(z-\frac{x}{4}\right)^2+\frac{x}{8}+\frac{4}{x}\\ \ge \frac{1}{2x}+\frac{x}{8}+\frac{4}{x}=\frac{9x}{2}+\frac{x}{8}\ge \frac{3}{2}, \end{aligned} \]

当且仅当\(\small (x,y,z)=(6,3,1)\)或\(\small (6,3,2)\)时等号成立;

(iv)如果\(\small x\equiv 3\pmod 4\)，那么

\[\small \begin{aligned} \frac{1}{x}\left(y-\frac{x}{2}\right)^2+\frac{2}{x}\left(z-\frac{x}{4}\right)^2+\frac{x}{8}+\frac{4}{x}\\ \ge \frac{1}{4x}+\frac{1}{8x}+\frac{x}{8}+\frac{4}{x}=\frac{35}{8x}+\frac{x}{8}\\ \ge \frac{3}{2}, \end{aligned} \]

当且仅当\(\small (x,y,z)=(7,3,2)\)或\(\small (7,4,2)\)时等号成立.

综上，\(\small \displaystyle \tau_0=\frac{3}{2}\).

(2)由(1)可知，此时\(\small (x,y,z)\)的所有正整数解为
$$\small (4,2,1),(8,4,2),(5,2,1),(5,3,1),(6,3,1),(6,3,2),(7,3,2),(7,4,2).$$