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

Discussion on Bangladesh National Math Camp
*Mahi*
Posts: 1175
Joined: Wed Dec 29, 2010 12:46 pm
Location: 23.786228,90.354974
Contact:

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

Edit your post, why don't you?

Use $L^AT_EX$, It makes our work a lot easier!

Rashik ctg
Posts: 8
Joined: Wed Sep 28, 2011 12:24 pm

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

1.7(i)

0 ≤ a ≤ b ≤ 1.
Does that imply,
a ≤ 1, b ≥ 0????
if so , we can't divide (b - a) by (1 - ab) . if (ab = 1), (1- ab = 0) ??
b ≥ a
=> b - a ≥ 0
=> (b - a) / 1 - ab ≥ 0 [ a≤b≤1 , ab≤1, 1 - ab not equal to 0] ****
again,
so, b≤1
=> b - a ≤ 1 [ a≤b≤1 ]****
=> (b - a)/ 1 - ab ≤ 1 [1-ab>0]

so 0 ≤(b - a)/ 1 - ab ≤ 1
help me. tell me if the **** noted assumptions r true or not.

nafistiham
Posts: 829
Joined: Mon Oct 17, 2011 3:56 pm
Location: 24.758613,90.400161
Contact:

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

it is said that $0\leq a\leq b\leq1$ so cant it be $a=b=0$?
at that case you cant divide anything by $1-ab$
$\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0$
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.

*Mahi*
Posts: 1175
Joined: Wed Dec 29, 2010 12:46 pm
Location: 23.786228,90.354974
Contact:

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

$ab=0$ directly implies an equality, so we may as well exclude that from our proof.

Use $L^AT_EX$, It makes our work a lot easier!

*Mahi*
Posts: 1175
Joined: Wed Dec 29, 2010 12:46 pm
Location: 23.786228,90.354974
Contact:

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

To exclude triviality from your solution, just state and prove those cases first, and then complete the proof.

Use $L^AT_EX$, It makes our work a lot easier!

Rashik ctg
Posts: 8
Joined: Wed Sep 28, 2011 12:24 pm

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

Prove which cases??? ****noted ones??

a≤b≤1
=> b-a≤1
=> (b-a)/1-ab ≤1 [1≥b≥a so ab≤1 1-ab ≥0]

does that satisfy??

*Mahi*
Posts: 1175
Joined: Wed Dec 29, 2010 12:46 pm
Location: 23.786228,90.354974
Contact:

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

Just to explain this, you can divide the thing in two cases , where $ab=1$ and where $ab \leq 1$. Then you can divide something by $1-ab$ when it is necessary.

Use $L^AT_EX$, It makes our work a lot easier!

Posts: 8
Joined: Mon Oct 24, 2011 12:13 am

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

Sad but true, আমি নিজেও সমাধান করতে পারিনি, উপরের ৪ টা সমধানের একটাও বুঝতে পারিনি!
Stay Hungry. Stay Foolish.

*Mahi*
Posts: 1175
Joined: Wed Dec 29, 2010 12:46 pm
Location: 23.786228,90.354974
Contact:

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

Please tell others in which part were you stuck. Then others can try and explain it to you.

Use $L^AT_EX$, It makes our work a lot easier!

rakeen
Posts: 384
Joined: Thu Dec 09, 2010 5:21 pm
Location: Dhaka

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

$0 \leq a$ and $0 \leq b$. substracting we get $b-a \geq 0$. again $a \leq 1$ and so is b. Similarly by substracting we get $b-a \leq 0$ !!! walla! so a=b. and $\frac{1}{4ab}$ is abviously greater than 1; so our inequality is proven. and for the first one, we merely just have to see that $a \leq b$ $\rightarrow ba^2 \leq ab^2$.
@Sourav Das how did you write $4ab(1-ab) \geq 4(ab^2 - a^2 b)$ ?
@*mahi* how can you say $4a + \frac{1}{ab} \geq 4b$ if $1 \geq b^{\frac{3}{2}}$ ?
@MATHPRITOM what a solution! but how would I know I've to take $(2a \sqrt{b} - 1)^2 \geq 0$ ? If we start from the backward it's not easy to say $\sqrt{b} \geq b \geq b^2$.

@*Mahi* "ab=0 directly implies an equality, so we may well exclude that from our proof"! why we are excluding that?the problem did asked ask to prove both inequality and equality. how can I show $\frac {b-a}{1-ab} \geq 1$ when ab=0 ? we can't divide with ab-1 right?
-rakeen
r@k€€/|/