Question

证明不等式lnx>2(x-1)/(x+1) (x>1) (用导数做）

Answer 1

lnx>2(x-1)/(x+1)

另f(x)=lnx-2(x-1)/(x+1)

求导 f'(x)=1/x-[(x+1)-(x-1)]/(x+1)^2

=1/x-2/(x+1)^2

=[(x+1)^2-2x]/(x+1)^2

=(x^2+1)/(x+1)^2>0

所以f(x)在定义域单调递增

且当x=1时

f(x)=0

所以当x>1时,f(x)>0

所以lnx-2(x-1)/(x+1)>0,即lnx>2(x-1)/(x+1)