解答：证明：∵a，b，c都是正数，∴a2b2+b2c2≥2ab2c，a2b2+c2a2≥2a2bc，c2a2+b2c2≥2abc2∴2（a2b2+b2c2+c2a2）≥2ab2c+2a2bc+2abc2∴a2b2+b2c2+c2a2≥ab2c+a2bc+abc2∴ a2b2+b2c2+c2a2 a+b+c ≥abc．
