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

得用a2+b2+c2+2ab+2ac+2bc=(a+b+c)2的变换来证明即可