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[不等式] 求助:一道不等式的证明

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敬畏数学 Posted 2024-7-13 14:10 |Read mode
$ x,y $均为正实数,求证:$ \frac{x^2}{x+2y} +\frac{4(y^2+2)}{xy+2}\geqslant 4$

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kuing Posted 2024-7-13 14:51
Last edited by kuing 2024-7-13 21:02给帖子贴上“二元不等式”的标签后,在自动生成的“相关帖子”一栏里马上就有一道看着就很像的题:
  • 正实数 $x,y$ , 求 $\frac{x^2}{xy+1}+\frac{y^2+2}{x+y}$ 的最小值

  • 1# 系数有点不平衡,把它平衡一下,作置换 `x\mapsto2x`,1# 就变成 \[\frac {x^2}{x+y}+\frac {y^2+2}{xy+1}\geqslant 2,\] 和那帖更像了,只是分母换了位置。方法自然可以照搬: \[\LHS\geqslant \frac {x^2}{\frac {x^2+1}2+\frac {y^2+1}2}+\frac {y^2+2}{\frac {x^2+y^2}2+1}=2.\]

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     Author| 敬畏数学 Posted 2024-7-19 15:51
    kuing 发表于 2024-7-13 14:51
    给帖子贴上“二元不等式”的标签后,在自动生成的“相关帖子”一栏里马上就有一道看着就很像的题:
    正实数  ...

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    山川浮云 Posted 2024-7-21 00:11
    Last edited by 山川浮云 2024-7-21 00:20练练输入操作

    \[\LHS\geqslant \frac{x^2}{\frac{x^2+4}{4}+\frac{4y^2+4}{4}}+\frac{4(y^2+2)}{\frac{x^2+4y^2}{4}+2}=4\]
    当且仅当\(x=2,y=1  \)时取得最小值.

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     Author| 敬畏数学 Posted 2024-7-21 09:47
    山川浮云 发表于 2024-7-21 00:11
    练练输入操作

    \[\LHS\geqslant \frac{x^2}{\frac{x^2+4}{4}+\frac{4y^2+4}{4}}+\frac{4(y^2+2)}{\frac{x ...

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    山川浮云 Posted 2024-7-21 17:22
    也跟一个,简单的类似题,已知\(x,y,z\)为正实数,求\(\frac{\sqrt {xy} +3\sqrt{yz}}{x+2y+z}\)的最大值.

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