已知abc∈R+ 求证a(b²+c²)+b(c²+a²)+c(a²+b²)≥6abc
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![已知abc∈R+ 求证a(b²+c²)+b(c²+a²)+c(a²+b²)≥6abc](/uploads/image/z/2506155-51-5.jpg?t=%E5%B7%B2%E7%9F%A5abc%E2%88%88R%2B+%E6%B1%82%E8%AF%81a%28b%26%23178%3B%2Bc%26%23178%3B%29%2Bb%28c%26%23178%3B%2Ba%26%23178%3B%29%2Bc%28a%26%23178%3B%2Bb%26%23178%3B%29%E2%89%A56abc)
已知abc∈R+ 求证a(b²+c²)+b(c²+a²)+c(a²+b²)≥6abc
已知abc∈R+ 求证a(b²+c²)+b(c²+a²)+c(a²+b²)≥6abc
已知abc∈R+ 求证a(b²+c²)+b(c²+a²)+c(a²+b²)≥6abc
∵a,b,c∈R+
∴a²+b²≥2ab
b²+c²≥2bc
a²+c²≥2ac
∴a(b²+c²)+b(c²+a²)+c(a²+b²)
≥a*2bc+b*2ac+c*2ab=6abc