## Exercise 1.7 (iii) [Section 1, BOMC-2011]

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Labib
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### Exercise 1.7 (iii) [Section 1, BOMC-2011]

The problem states,

If $a,b$ are real numbers such that $0 \leq a \leq b \leq 1$,
Prove that,
$0 \leq ab^2-a^2b \leq \frac{1}{4}$.
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Labib
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### Re: Exercise 1.7 (iii) [Section 1, BOMC-2011]

Here's how far I managed to take it...
Need further help or other sort of hints.
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sourav das
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### Re: Exercise 1.7 (iii) [Section 1, BOMC-2011]

Note that $b^2 \leq b$ . Now $\{ ab + ( 1-ab )\}^2 \geq 4ab(1-ab) \geq 4(ab^2 - a^2b)$
Done. Equality holds when $a(a+.5)=1$ & $b=a+.5$
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sourav das
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### Re: Exercise 1.7 (iii) [Section 1, BOMC-2011]

By the way, I couldn't find out the problem in the book.
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

Labib
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### Re: Exercise 1.7 (iii) [Section 1, BOMC-2011]

Please Install $L^AT_EX$ fonts in your PC for better looking equations,
Learn how to write equations, and don't forget to read Forum Guide and Rules.

"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes

*Mahi*
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### Re: Exercise 1.7 (iii) [Section 1, BOMC-2011]

It can also be solved by quick application of AM-GM inequality.
Now, $4a+\frac 1 {ab} \geq 2\sqrt{4a \cdot \frac 1 {ab}} = 4 \sqrt {\frac 1 b}$
As $1 \geq b^{\frac 3 2}$ so that inequality gives us $4a+\frac 1 {ab} \geq 4b$ , or ,$4a^2b+1 \geq 4ab^2$.
Proved!

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otpid
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### Re: Exercise 1.7 (iii) [Section 1, BOMC-2011]

the problems in the downloaded copy are not matching with the printed one. What to do now??

Dipto

MATHPRITOM
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### Re: Exercise 1.7 (iii) [Section 1, BOMC-2011]

It is enough to prove 4$a^{2}$ b+1$\ge$ $4ab^{2}$.
now,$(2a\sqrt{b} -1)^2$ $\ge$ 0.
or,4$a^2$b+1 $\ge$ 4a$\sqrt{b}$.
now,$1\ge b;$
or, $\sqrt{b} \ge b\ge b^{2}$
so,4$a^2$b+1 $\ge$ 4a$\sqrt{b} \ge 4ab^{2}$.
done.
Last edited by MATHPRITOM on Wed Oct 26, 2011 6:04 pm, edited 1 time in total.

Labib
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### Re: Exercise 1.7 (iii) [Section 1, BOMC-2011]

MATHPRITOM wrote:now,$b\ge1;$
I think you meant to say $b\leq 1$.

BTW, Really nice solutions, you three...
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MATHPRITOM
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### Re: Exercise 1.7 (iii) [Section 1, BOMC-2011]

ya,labib.i wrote a wrong here......i made the mistake at the time of typing.thanks.