Tighter logarithmic inequality There is a well-known lower bound for x log((1+x)/x)>=(x/(1+x)) for x>=0. I know a tighter lower bound on the same domain x\log((1+x)/x)>=(2x)/(1+2x)>=x/(1+x). It can be proved by Jensen's inequality. Do you know something tighter?

spoofing44

spoofing44

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2022-08-20

Tighter logarithmic inequality
There is a well-known lower bound for
x log 1 + x x x 1 + x
for x 0. I know a tighter lower bound on the same domain
x log 1 + x x 2 x 1 + 2 x x 1 + x .
It can be proved by Jensen's inequality. Do you know something tighter?

Answer & Explanation

Dillan Brock

Dillan Brock

Beginner2022-08-21Added 8 answers

If any tighter lower bound will suffice, here you go with a ratio of quadratics:
x log x + 1 x 3 x ( 2 x + 1 ) 6 x 2 + 6 x + 1 2 x 2 x + 1
Proving the rightmost inequality is easy with cross multiplication to get x 0. For the left inequality, you need to show for x > 0
f ( x ) = log ( 1 + 1 x ) 6 x + 3 6 x 2 + 6 x + 1 0
which is evident from f ( x ) = 1 x ( x + 1 ) ( 6 x 2 + 6 x + 1 ) 2 < 0 and lim x f ( x ) = 0
P.S. You can keep getting better approximations using larger polynomials p ( x ) q ( x ) - one way is to ensure that this matches in limits ( x 0 , x ) with your LHS, then get as many terms of the Taylor series (or as x ) to match as you want. You may also want to google or read up on Pade approximants.

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