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]

Unread post by Labib » Tue Oct 25, 2011 9:50 pm

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

Unread post by Labib » Tue Oct 25, 2011 9:55 pm

Here's how far I managed to take it...
Let $(b-a)=k$, where $k$ is another real number.
Then I simplified the problem and deduced that it would be enough to prove that $\frac{1}{2a}>(2k)^2$.
Need further help or other sort of hints.
:ugeek:
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Re: Exercise 1.7 (iii) [Section 1, BOMC-2011]

Unread post by sourav das » Tue Oct 25, 2011 10:20 pm

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 $a(a+.5)=1$ & $b=a+.5$
You spin my head right round right round,
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Re: Exercise 1.7 (iii) [Section 1, BOMC-2011]

Unread post by sourav das » Tue Oct 25, 2011 10:22 pm

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

Unread post by Labib » Tue Oct 25, 2011 10:25 pm

Download the book's year 2009 edition...
Your one's year 2005 edition...
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Re: Exercise 1.7 (iii) [Section 1, BOMC-2011]

Unread post by *Mahi* » Tue Oct 25, 2011 10:52 pm

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

Unread post by otpid » Wed Oct 26, 2011 12:04 pm

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

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

Unread post by MATHPRITOM » Wed Oct 26, 2011 12:26 pm

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.

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

Unread post by Labib » Wed Oct 26, 2011 1:24 pm

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

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

Unread post by MATHPRITOM » Wed Oct 26, 2011 2:01 pm

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

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