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[不等式] 一眼看尽 $\frac 1x+\frac 2y$ 的最小值

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isee Posted 2022-3-29 19:41 |Read mode
源自知乎提问



:已知正实数满足 $x+3y=2$,则 $\frac 1x+\frac 2y$ 的最小值为_____.

解答先不上,哈哈哈~

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战巡 Posted 2022-3-29 20:54
回复 1# isee


挺无聊的....那个解答还非要绕一个大圈

\[2(\frac{1}{x}+\frac{2}{y})=(x+3y)(\frac{1}{x}+\frac{2}{y})=7+\frac{2x}{y}+\frac{3y}{x}\ge 7+2\sqrt{\frac{2x}{y}\cdot\frac{3y}{x}}=7+2\sqrt{6}\]
\[\frac{1}{x}+\frac{2}{y}\ge\frac{7}{2}+\sqrt{6}\]
当$\frac{2x}{y}=\frac{3y}{x}$时等号成立

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 Author| isee Posted 2022-3-29 21:21
回复 2# 战巡

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