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[不等式] 已知 a,b,c,d>0,证明 (a+b+c+d)/(abcd)<=1/a^3+1/b^3+1/c^3+1/d^3

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TSC999 posted 2022-6-14 21:36 |Read mode
\(a,b,c,d>0\),证明 \(\frac{a+b+c+d}{abcd}\leqslant \frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{1}{d^3}\)

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kuing posted 2022-6-14 21:38
太简单了吧,作代换 a->1/x 等,就是 xyz+yzw+zwx+wxy<=x^3+y^3+z^3+w^3

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original poster TSC999 posted 2022-6-15 10:51
按 K 先生的变换法证:

变换.png

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