Penchick Online Mathematical Olympiad, Qualifying Test 1, III.4

原题链接：https://artofproblemsolving.com/community/c4h3687790_1st_penchick_online_mathematical_olympiad_qualifying_test_1_iii4

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解答：

令$p = 3a + b$ ， $q = 2a + 3b$. （统一分母）
整理得：
$$a = \frac{3p - q}{7}, b = \frac{2p - 3q}{7}.$$代入原式
\begin{align*} 9b - a = \frac{28q - 21p}{7} = 4q - 3p;\\ 4a - b = \frac{14p - 7q}{7} = 2p - q. \end{align*}
$$\frac{9b - a}{3a + b} + \frac{4a - b}{2a + 3b} = 4(\frac{q}{p}) + 2(\frac{p}{q}) - 4.$$
$$4(\frac{q}{p}) + 2(\frac{p}{q}) \geq 2\sqrt{8 \cdot \frac{q}{p} \cdot \frac{p}{q}} = 2\sqrt{8} = 4\sqrt{2}.$$

均值不等式(AM-GM)得到最小 $4\sqrt{2} - 4$, 答案为$\boxed{424}$.

posted @ 2025-10-07 23:08  JMCE  阅读(25)  评论(2)    收藏  举报