Question

prove that x square log(1/x)​

Answer 1

7

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Intuition behind logarithm inequality: 1−1x≤logx≤x−1 (4 answers)

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I want to show that

x1+x<log(1+x)<x

for all x>0 using the mean value theorem. I tried to prove the two inequalities separately.

x1+x<log(1+x)⇔x1+x−log(1+x)<0

Let

f(x)=x1+x−log(1+x).

Since

f(0)=0

and

f′(x)=1(1+x)2−11+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 x0 between a and x with

f′(x0)=f(x)−f(a)x−a⇔1x0+1=log(x+1)x.

Since

x0>0⇒1x0+1<1.

⇒1>1x0+1=log(x+1)x⇒x>log(x+1)

calculus real-analysis inequality logarithms

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edited Jun 17 '14 at 16:34

Martin Sleziak

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asked Jan 26 '14 at 23:28

fear.xD

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How did you arrive at that first equivalence, x1+x<log(1+x)⟺−log(1+x)x1+x<0? Also, you've calculated f′(x) incorrectly. – Amit Kumar Gupta Jan 26 '14 at 23:36

By subtracting it. Wolfram Alpha says, my derivation is correct... – fear.xD Jan 26 '14 at 23:40

Hint: If you subtract, there should be a plus sign to indicate addition. The way you have it now looks like they're being multiplied. – MCT Jan 26 '14 at 23:41  

I've just copied it wrong, but used it correct in my derivation.. – fear.xD Jan 26 '14 at 23:43

I know; it was tongue in cheek. – MCT Jan 26 '14 at 23:44