#int_1^2((e^x-lnx)/x^2-1)dx>0# #<=>#
#int_1^2((e^x-lnx)/x^2)dx>int_1^2(1)dx# #<=>#
#int_1^2((e^x-lnx)/x^2)dx>[x]_1^2# #<=># #<=>#
#int_1^2((e^x-lnx)/x^2)dx>2-1# #<=>#
#int_1^2((e^x-lnx)/x^2)dx>1#
We need to prove that
#int_1^2((e^x-lnx)/x^2)dx>1#
Consider a function #f(x)=e^x-lnx# , #x>0#
From the graph of #C_f# we can notice that for #x>0#
we have #e^x-lnx>2#
Explanation:
#f(x)=e^x-lnx# , #x##in##[1/2,1]#
#f'(x)=e^x-1/x#
#f'(1/2)=sqrte-2<0#
#f'(1)=e-1>0#
According to Bolzano (Intermediate Value) Theorem we have #f'(x_0)=0# #<=># #e^(x_0)-1/x_0=0# #<=>#
#e^(x_0)=1/x_0# #<=># #x_0=-lnx_0#
The vertical distance is between #e^x# and #lnx# is minimum when #f(x_0)=e^(x_0)-lnx_0=x_0+1/x_0#
We need to show that #f(x)>2# , #AAx##>0#
#f(x)>2# #<=># #x_0+1/x_0>2# #<=>#
#x_0^2-2x_0+1>0# #<=># #(x_0-1)^2>0# #-># true for #x>0#
graph{e^x-lnx [-6.96, 7.09, -1.6, 5.42]}
#(e^x-lnx)/x^2>2/x^2#
#int_1^2((e^x-lnx)/x^2)dx>int_1^2(2/x^2)dx# #<=>#
#int_1^2((e^x-lnx)/x^2)dx>[-2/x]_1^2# #<=>#
#int_1^2((e^x-lnx)/x^2)dx>##-1+2# #<=>#
#int_1^2((e^x-lnx)/x^2)dx>1# #<=>#
#int_1^2((e^x-lnx)/x^2-1)dx>0#
