(2024年中國東南地區數學奧林匹克競賽第1題)若實數\(\small \tau\)滿足：對任意正整數\(\small x,y,z\)，均有
\[\small \small x^2+2y^2+4z^2+8\ge 2x(y+z+\tau),
\]
則稱\(\small \tau\)為“平生”數.記最大的平生數為\(\small \tau_0\).
(1)求\(\small \tau_0\)的值;
(2)求方程\(\small x^2+2y^2+4z^2+8= 2x(y+z+\tau_0)\)的所有正整數解\(\small (x,y,z)\).
解：(1)原不等式可等價轉換為
\[\small \frac{x}{2}+\frac{y^2}{x}+\frac{2z^2}{x}+\frac{4}{x}-y-z\ge \tau,
\]
注意到
\[\small \text{LHS}=\frac{1}{x}\left(y-\frac{x}{2}\right)^2+\frac{2}{x}\left(z-\frac{x}{4}\right)^2+\frac{x}{8}+\frac{4}{x}.
\]
因為\(\small x,y,z\)是正整數，所以需要對\(x\)進行分類討論：
(i)如果\(\small x\equiv 0\pmod 4\)，那麼
\[\small
\begin{aligned}
\frac{1}{x}\left(y-\frac{x}{2}\right)^2+\frac{2}{x}\left(z-\frac{x}{4}\right)^2+\frac{x}{8}+\frac{4}{x}\\
\ge \frac{x}{8}+\frac{4}{x}\ge \frac{3}{2},
\end{aligned}
\]
當且僅當\(\small (x,y,z)=(4,2,1)\)或\(\small (8,4,2)\)時等號成立;
(ii)如果\(\small x\equiv 1\pmod 4\)，那麼
\[\small
\begin{aligned}
\frac{1}{x}\left(y-\frac{x}{2}\right)^2+\frac{2}{x}\left(z-\frac{x}{4}\right)^2+\frac{x}{8}+\frac{4}{x}\\
\ge \frac{1}{4x}+\frac{1}{8x}+\frac{x}{8}+\frac{4}{x}=\frac{35}{8x}+\frac{x}{8} \\
\ge \frac{3}{2},
\end{aligned}
\]
當且僅當\(\small (x,y,z)=(5,2,1)\)或\(\small (5,3,1)\)時等號成立;
(iii)如果\(\small x\equiv 2\pmod 4\)，那麼
\[\small
\begin{aligned}
\frac{1}{x}\left(y-\frac{x}{2}\right)^2+\frac{2}{x}\left(z-\frac{x}{4}\right)^2+\frac{x}{8}+\frac{4}{x}\\
\ge \frac{1}{2x}+\frac{x}{8}+\frac{4}{x}=\frac{9x}{2}+\frac{x}{8}\ge \frac{3}{2},
\end{aligned}
\]
當且僅當\(\small (x,y,z)=(6,3,1)\)或\(\small (6,3,2)\)時等號成立;
(iv)如果\(\small x\equiv 3\pmod 4\)，那麼
\[\small
\begin{aligned}
\frac{1}{x}\left(y-\frac{x}{2}\right)^2+\frac{2}{x}\left(z-\frac{x}{4}\right)^2+\frac{x}{8}+\frac{4}{x}\\
\ge \frac{1}{4x}+\frac{1}{8x}+\frac{x}{8}+\frac{4}{x}=\frac{35}{8x}+\frac{x}{8}\\
\ge \frac{3}{2},
\end{aligned}
\]
當且僅當\(\small (x,y,z)=(7,3,2)\)或\(\small (7,4,2)\)時等號成立.
綜上，\(\small \displaystyle \tau_0=\frac{3}{2}\).
(2)由(1)可知，此時\(\small (x,y,z)\)的所有正整數解為
$$\small (4,2,1),(8,4,2),(5,2,1),(5,3,1),(6,3,1),(6,3,2),(7,3,2),(7,4,2).$$