已知abc属于R+求证 1.(a+b+c)(a^2+b^2+c^2)≥9abc (2).2.(a/b+b/c+c/a)(b/a+c/b+a/c)≥9
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![已知abc属于R+求证 1.(a+b+c)(a^2+b^2+c^2)≥9abc (2).2.(a/b+b/c+c/a)(b/a+c/b+a/c)≥9](/uploads/image/z/2736785-65-5.jpg?t=%E5%B7%B2%E7%9F%A5abc%E5%B1%9E%E4%BA%8ER%2B%E6%B1%82%E8%AF%81+1.%28a%2Bb%2Bc%29%28a%5E2%2Bb%5E2%2Bc%5E2%29%E2%89%A59abc+%282%29.2.%28a%2Fb%2Bb%2Fc%2Bc%2Fa%29%28b%2Fa%2Bc%2Fb%2Ba%2Fc%29%E2%89%A59)
已知abc属于R+求证 1.(a+b+c)(a^2+b^2+c^2)≥9abc (2).2.(a/b+b/c+c/a)(b/a+c/b+a/c)≥9
已知abc属于R+求证 1.(a+b+c)(a^2+b^2+c^2)≥9abc (2).
2.(a/b+b/c+c/a)(b/a+c/b+a/c)≥9
已知abc属于R+求证 1.(a+b+c)(a^2+b^2+c^2)≥9abc (2).2.(a/b+b/c+c/a)(b/a+c/b+a/c)≥9
abc属于R+
由均值不等式
a+b+c>=3(abc)的立方根
a^2+b^2+c^2>=3(a^2b^2c^2)的立方根
所以(a+b+c)(a^2+b^2+c^2)>=9*(a^3b^3c^3)的立方根
即(a+b+c)(a^2+b^2+c^2)>=9abc
同理
a/b+b/c+c/a>=3(a/b*b/c*c/a)的立方根=3
b/a+c/b+a/c>=3(b/a*c/b*a/c)的立方根=3
所以(a/b+b/c+c/a)(b/a+c/b+a/c)>=9