Prove that lnx<=x−1 I need help with this proof for my real analysis class. it is on the practice sheets and we do NOT get an answer. I proved ln(x)<x−1 for all x>1 by contradiction but cannot do this one. Prove that ln(x)<=x−1 for all x>0.i believe you need to use MVT, I cannot use the famous inequality e^x>x+1 for all x>0.

djedgaravilafo

djedgaravilafo

Open question

2022-08-22

Prove that ln x x 1
I need help with this proof for my real analysis class. it is on the practice sheets and we do NOT get an answer. I proved ln ( x ) < x 1 for all x > 1 by contradiction but cannot do this one.
Prove that ln ( x ) x 1 for all x > 0
i believe you need to use MVT, I cannot use the famous inequality e x > x + 1 for all x > 0

Answer & Explanation

Coraducci0d

Coraducci0d

Beginner2022-08-23Added 8 answers

Apply what you've already proven using ln ( 1 x ) = ln ( x )
Giltrapsx

Giltrapsx

Beginner2022-08-24Added 2 answers

Note that, for y > 1
1 y d t t 2 1 y d t t 1 y d t
gives
(1) 1 1 y log y y 1
with equality iff y = 1. If 0 < y < 1, write y = 1 y and use the above to get
1 y log 1 y 1 y 1
and multiply by 1 to get ( 1 ) is true also in this case.

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