设a,b,c是实数,求证：a^2b^2+b^2c^2+c^2a^2>=abc(a+b+c)
人气：206 ℃ 时间：2019-10-29 06:19:30
解答
a,b,c>0.因为a^2b^2+b^2c^2=b^2(a^2+c^2)>=2acb^2同理有b^2c^2+c^2a^2>=2abc^2c^2a^2+a^2b^2>=2bca^2故三式相加得2(a^2b^2+b^2c^2+c^2a^2)>=2(abc^2+acb^2+bca^2)=2abc(a+b+c)a^2b^2+b^2c^2+c^2a^2>=abc(a+b+c)...
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