Exercise 1.7 (iii) [Section 1, BOMC2011]
Exercise 1.7 (iii) [Section 1, BOMC2011]
The problem states,
If $a,b$ are real numbers such that $0 \leq a \leq b \leq 1$,
Prove that,
$0 \leq ab^2a^2b \leq \frac{1}{4}$.
If $a,b$ are real numbers such that $0 \leq a \leq b \leq 1$,
Prove that,
$0 \leq ab^2a^2b \leq \frac{1}{4}$.
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
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
Re: Exercise 1.7 (iii) [Section 1, BOMC2011]
Here's how far I managed to take it...
Need further help or other sort of hints.
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
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

 Posts: 461
 Joined: Wed Dec 15, 2010 10:05 am
 Location: Dhaka
 Contact:
Re: Exercise 1.7 (iii) [Section 1, BOMC2011]
Note that $b^2 \leq b$ . Now $\{ ab + ( 1ab )\}^2 \geq 4ab(1ab) \geq 4(ab^2  a^2b)$
Done. Equality holds when $a(a+.5)=1$ & $b=a+.5$
Done. Equality holds when $a(a+.5)=1$ & $b=a+.5$
You spin my head right round right round,
When you go down, when you go down down......($from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......($from$ "$THE$ $UGLY$ $TRUTH$" )

 Posts: 461
 Joined: Wed Dec 15, 2010 10:05 am
 Location: Dhaka
 Contact:
Re: Exercise 1.7 (iii) [Section 1, BOMC2011]
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......($from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......($from$ "$THE$ $UGLY$ $TRUTH$" )
Re: Exercise 1.7 (iii) [Section 1, BOMC2011]
Download the book's year 2009 edition...
Your one's year 2005 edition...
Your one's year 2005 edition...
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
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
Re: Exercise 1.7 (iii) [Section 1, BOMC2011]
It can also be solved by quick application of AMGM 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!
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!
Please read Forum Guide and Rules before you post.
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah  Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah  Mahi
Re: Exercise 1.7 (iii) [Section 1, BOMC2011]
the problems in the downloaded copy are not matching with the printed one. What to do now??
Dipto
Dipto

 Posts: 190
 Joined: Sat Apr 23, 2011 8:55 am
 Location: Khulna
Re: Exercise 1.7 (iii) [Section 1, BOMC2011]
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.
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.
Re: Exercise 1.7 (iii) [Section 1, BOMC2011]
I think you meant to say $b\leq 1$.MATHPRITOM wrote:now,$b\ge1;$
BTW, Really nice solutions, you three...
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
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

 Posts: 190
 Joined: Sat Apr 23, 2011 8:55 am
 Location: Khulna
Re: Exercise 1.7 (iii) [Section 1, BOMC2011]
ya,labib.i wrote a wrong here......i made the mistake at the time of typing.thanks.