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[不等式] 三元求值

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APPSYZY Posted 2020-12-6 01:55 |Read mode

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Author| APPSYZY Posted 2020-12-6 02:16

如果a+b+c=0的话就容易了

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kuing Posted 2020-12-6 02:49

2(ab+bc+ca) = (a+b+c)^2-(a^2+b^2+c^2) = 0
a^5+b^5+c^5 = (a+b+c)^5-5(a+b+c)^3(ab+bc+ca)+5abc(a+b+c)^2+5(a+b+c)(ab+bc+ca)^2-5abc(ab+bc+ca) = 1+5abc
所求值不确定啊

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