设a,b,c是实数,求证：a^2b^2+b^2c^2+c^2a^2>=abc(a+b+c)

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)...