Let x > 0, prove x^2 + 1 > lnx?

2 Answers
Jun 22, 2018

Two cases: #0 < x < 1 # and #x ge 1#

For #x<1# the left side # x^2 +1# is positive and the right side #ln x# is negative, so the statement is true.

For # x ge 1# consider

#f(x) = x^2 + 1 - ln x#

#f'(x) = 2x - 1/x #

For all #x ge 1# we see #f'(x) > 0# so #f# is increasing over the domain #x ge 1.# Since #f(1)=2>0# and #f# is increasing, we have #f(x)>0# for all #x ge 1# or

# x^2 +1 > ln x#

We've shown both cases, so this is true for all #x >0.#

Jun 22, 2018

As #d/dx lnx=1/x# grows much slower than #d/dx (x^2+1)=2x#, it follows that #x^2+1>lnx#

Explanation:

For any number #0 < x <= 1# we have #lnx=<0#, so it's quite obvious that in this interval #x^2+1>lnx# in this interval.

#d/dx lnx=1/x# (I won't prove it here - you can find the proof on the net.
As #d/dx (x^2+1)=2x# we can see that #x^2+1# increases much faster than #lnx#. It, therefore, follows that #x^2+1>lnx#

Graphically it is quite clear:

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