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1a3(b+c)+1b3(c+a)+1c3(a+b)≥AB\frac{1}{a^{3}(b+c)} + \frac{1}{b^{3}(c+a)} + \frac{1}{c^{3}(a+b)} \geq \frac{A}{B}a3(b+c)1+b3(c+a)1+c3(a+b)1≥BA
The inequality above, where AAA and BBB are coprime, holds true for abc=1abc = 1abc=1 and a,b,c>0a, b, c > 0a,b,c>0. Find A+BA+BA+B.
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