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DIAMMIBENVERMk1 2022-07-03 Answered
Showing x 1 + x < log ( 1 + x ) < x for all x > 0 using the mean value theorem
I want to show that
x 1 + x < log ( 1 + x ) < x
for all x > 0 using the mean value theorem. I tried to prove the two inequalities separately.
x 1 + x < log ( 1 + x ) x 1 + x log ( 1 + x ) < 0
Let
f ( x ) = x 1 + x log ( 1 + x ) .
Since
f ( 0 ) = 0
and
f ( x ) = 1 ( 1 + x ) 2 1 1 + x < 0
for all x > 0, f ( x ) < 0 for all x > 0. Is this correct so far?
I go on with the second part: Let
f ( x ) = log ( x + 1 ). Choose a = 0 and x > 0 so that there is, according to the mean value theorem, an x 0 between a and x with
f ( x 0 ) = f ( x ) f ( a ) x a 1 x 0 + 1 = log ( x + 1 ) x
Since
x 0 > 0 1 x 0 + 1 < 1.
1 > 1 x 0 + 1 = log ( x + 1 ) x x > log ( x + 1 )
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Answers (2)

razdiralem
Answered 2022-07-04 Author has 15 answers
By Definition of log(which already a kind of mean value theorem ) we have For any x > 0 We have
x x + 1 = 0 x d t x + 1 0 x d t t + 1 = ln ( x + 1 ) = 0 x d t t + 1 0 x d t 1 = x
Thus,
x x + 1 ln ( x + 1 ) x
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Lena Bell
Answered 2022-07-05 Author has 4 answers
Substitute x = y 1 to get
1 y 1 < log y < y 1
for all y > 1. Now note
1 y t 2 d t < 1 y t 1 d t < 1 y d t
for y > 1
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