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我的推导过程 1/a = (1/b+1/c) / (1 - (1/b) * (1/c))
1/a = (b+c)/bc / (1-1/bc)
a = (bc-1) / (b+c)
令 m = b+c
则 c = m - b;
m = (b^2+1)/(b-a)
令 i = b-a
则 m = i + 2a + (a^2+1)/i //最后表达式
i+ (a^2+1)/i >= 2*sqrt(a^2+1)
所以: i >= sqrt(a^2+1) 或 i<= sqrt(a^2+1)
则 m >= 2*sqrt(a^2+1) + 2a
所以可以令 i 从 a 开始循环 i--;
直到 (a^2+1)%i == 0 再求出m即可
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