证明:当x>0时,arctanx+1/x>π/2.

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问题 证明:当x>0时,arctanx+1/x>π/2.

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答案令f =arctanx+1/x,因为f’(x)=1/(1+x2)-1/x2<0(x>0),所以f(x)在(0,+∞)内单调递减,又因为[*],所以f(x)>π/2,即arctanx+1/x>π/2.

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