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[不等式] 求三元代数式的最小值

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lemondian posted 2021-3-25 08:31 |Read mode
已知$a,b,c$为正实数,求$\dfrac{(a+b)^3+a^3}{(a+b)^3+b^3}+\dfrac{(b+c)^3+b^3}{(b+c)^3+c^3}+\dfrac{(c+a)^3+c^3}{(c+a)^3+a^3}$的最小值。

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kuing posted 2021-3-25 23:29
看着吓人,原来只要因式分解 `(a+b)^3+a^3=(2a+b)(a^2+ab+b^2)`,分母同理,立马简单起来,原式变成
\[\frac{2a+b}{a+2b}+\frac{2b+c}{b+2c}+\frac{2c+a}{c+2a},\]也就是
\[\frac32+\frac32\sum\frac a{a+2b},\]CS 即得最小值就是 `3`。

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