Solutions to 2017 Chinese Mathematical Olympiad(CMO)
不是完整的试题解答, 仅仅在关隘的地点聊一聊. 付云皓的题 3 的解 记 \(a=[nq^{\frac13}]\), \(b=[nq^{\frac23}]\), \(c=nq\). 然后 \begin{equation}\Big(c-aq^{\frac23}\Big)^2+\Big(c-bq^{\frac13}\Big)^2+\Big(aq^{\frac23}-bq^{\frac13}\Big)^2\\=\frac{2(a^3q^2+b^3q+c^3-3abcq)}{aq^{\frac23}+bq^{\frac13}+c}\\\geqslant\frac2{3c},\end{equation} 最后的不等式是因为 \(a^3q^2+b^3q+c^3\gt 3abcq\), 并且 \(c\geqslant aq^{\frac23}\), \(c\geqslant bq^{\frac13}\). 然后, 因为 \(c-aq^{\frac23}\geqslant0\), \(c- bq^{\frac13}\geqslant0\), 以及 \(aq^{\frac23}-bq^{\frac13}=-\Big((c-aq^{\frac23})-(c-bq^{\frac13})\Big)\), 得到 \begin{equation}\Big(c-aq^{\frac23}\Big)^2+\Big(c-bq^{\frac13}\Big)^2\leqslant\Big(2c-aq^{\frac23}-bq^{\frac13}\Big)^2,\end{equation} 与 \begin{equation}\Big(aq^{\frac23}-bq^{\frac13}\Big)^2\leqslant\Big(2c-aq^{\frac23}-bq^{\frac13}\Big)^2.\end{equation} 现在, \((1)\) …
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