证明不等式:a4+b4+c4≥a2b2+b2c2+c2a2≥abc(a+b+c)
问题描述:
证明不等式:a4+b4+c4≥a2b2+b2c2+c2a2≥abc(a+b+c)
答
证明:∵a4+b4≥2a2b2,b4+c4≥2b2c2,c4+a4≥2a2c2∴2(a4+b4+c4)≥2(a2b2+b2c2+a2c2)即a4+b4+c4≥a2b2+b2c2+a2c2又a2b2+b2c2≥2ab2c;b2c2+a2c2≥2abc2;a2b2+a2c2≥2a2bc∴2(a2b2+b2c2+a2c2)≥2(a2bc+ab2c+...