设a,b,c为正实数,求证：a^4+b^4+c^4>=a^2b^2+b^2c^2+c^2a^2>=abc(a+b+c).

2a4+2b4+2c4=(a4+b4)+(a4+c4)+(b4+c4)>=2a2b2+2a2c2+2b2c2即a^4+b^4+c^4>=a^2b^2+b^2c^2+c^2a^2同理,2(a^2b^2+b^2c^2+c^2a^2)＝2(ab)2+2(bc)2+2(ac)2=[(ab)2+(bc)2]+[(bc)2+(ac)2]+[(ab)2+(ac)2]>=2acb2+2abc2+2bca2=2abc(a+b+c)即a^2b^2+b^2c^2+c^2a^2>=abc(a+b+c)没有^号,将就着看吧