(1)设函数f(x)=xlnx+(1-x)ln(1-x)(0<x<1),求f(x)的最小值; (2)设正数 p 1 , p 2 , p 3 ,…, p 2 n 满足 p 1 + p 2 + p 3 +…+ p 2 n =1,求证: p 1 ln p 1 + p 2 ln p 2 + p 3 ln p 3 +…+ p 2 n ln p 2 n ≥-n.
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