题目

证明不等式:a,b,c∈R,a4+b4+c4≥abc(a+b+c). 答案:证明 ∵a4+b4≥2a2b2,b4+c4≥2b2c2, c4+a4≥2c2a2, ∴2(a4+b4+c4)≥2(a2b2+b2c2+c2a2) 即a4+b4+c4≥a2b2+b2c2+c2a2. 又a2b2+b2c2≥2ab2c,b2c2+c2a2≥2abc2, c2a2+a2b2≥2a2bc. ∴2(a2b2+b2c2+c2a2)≥2(ab2c+abc2+a2bc), 即a2b2+b2c2+c2a2≥abc(a+b+c). ∴a4+b4+c4≥abc(a+b+c).
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