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因為 AM >= GM, 所以
(x + y)/2 >= (xy)^(1/2) 及 (x + y + z)/3 >= (xyz)^(1/3)
即 (x + y) >= 2*(xy)^(1/2) 及 (x + y + z) >= 3*(xyz)^(1/3)

(a/b + b/c)^2014 + (b/c + c/a)^2014 + (c/a + a/b)^2014
>= 2^2014 * [(a/b)(b/c)]^(2014/2) + 2^2014 * [(b/c)(c/a)]^(2014/2) + 2^2014 * [(c/a)(a/b)]^(2014/2)
 = 2^2014 * (a/c)^1007 + 2^2014 * (b/a)^1007 + 2^2014 * (c/b)^1007
 = 2^2014 * [(a/c)^1007 + (b/a)^1007 + (b/a)^1007]
>= 2^2014 * 3 * [(a/c)^1007 * (b/a)^1007 * (b/a)^1007]^(1/3)
 = 3 * 2^2014

所以 (a/b + b/c)^2014 + (b/c + c/a)^2014 + (c/a + a/b)^2014 >= 3 * 2^2014

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