证明:当x >0时,arctan x+1/x>π/2

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证明:当x>0时,arctanx+1/x>π/2证明:当x>0时,arctanx+1/x>π/2证明:当x>0时,arctanx+1/x>π/2f''(x)=1/(1+x^2)-1/x^2=-1/(1+

证明:当x >0时,arctan x+1/x>π/2
证明:当x >0时,arctan x+1/x>π/2

证明:当x >0时,arctan x+1/x>π/2
f'(x)=1/(1+x^2)-1/x^2=-1/(1+x^2)x^2<0
f(x)在x>0上单调递减
lim(x趋近于正无穷)f(x)=π/2+0=π/2
∴arctan x+1/x>π/2

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