# Inequality Proof (fitzpatrick review 8, q5d) (1 Viewer)

#### jaccs

##### New Member
Trying to prove part d of the inequality proof from fitz chapter 8 review, question 5

parts a-c are fine

part d am stuck

from previous parts got as far as

$\bg_white 2{I_1} > \frac{{3{{(\ln 2)}^2}}}{2}$

and I1=2ln2-1

but can't finish the proof

did try replacing (ln 2)2 with just ln 2 as (ln 2)2 > ln2

any help greatly appreciated

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#### integral95

##### Well-Known Member
Trying to prove part d of the inequality proof from fitz chapter 8 review, question 5

parts a-c are fine

part d am stuck

from previous parts got as far as

$\bg_white 2{I_1} > \frac{{3{{(\ln 2)}^2}}}{2}$

and I1=2ln2-1

but can't finish the proof

did try replacing (ln 2)2 with just ln 2 as (ln 2)2 > ln2

any help greatly appreciated
That's not even true since you're squaring a number less than 1

$\bg_white 0 (ln(2))^2$

$\bg_white 2ln(2) > (ln(2))^2 \Rightarrow 2 > ln(2)$

At the moment I still can't figure the lower bound either.

#### InteGrand

##### Well-Known Member
Trying to prove part d of the inequality proof from fitz chapter 8 review, question 5

parts a-c are fine

part d am stuck

from previous parts got as far as

$\bg_white 2{I_1} > \frac{{3{{(\ln 2)}^2}}}{2}$

and I1=2ln2-1

but can't finish the proof

did try replacing (ln 2)2 with just ln 2 as (ln 2)2 > ln2

any help greatly appreciated
HINT: Using what you have, obtain a quadratic inequality for ln 2 and solve the inequality. This will get you (both of) the desired bounds.

#### jaccs

##### New Member
Thank you Integrand, that worked beautifully. You are genius.

Thanks integral95, that squaring was really careless of me, didn't even occur to me that it was less than 1, hence that inequality wouldn't hold, was just so desperate to get rid of the square... argghhh. will watch out for that in future.

thank you both, really appreciate it.