E-StoreHome » Ask & Discuss » Mathematics. » Algebra « Back to Discussion
Algebra
3 Mar 2008 08:18:47 IST
0 People liked this
21
1144
Comments (21)
3 Mar 2008 12:45:37 IST
Like
0 people liked this
@Decoder: Was that supposed to be a vedic chant or something. Lots of mumbling which didnt make any sense
Incidentally, I know the solution. You cant impress me by dropping terms like AM-GM inequality and so on. So, show me some solid working fella.
Also, if you happen to know the upper bound for this expression, pls go ahead, bcos that is the next qn I have in mind.
3 Mar 2008 12:52:59 IST
Like
0 people liked this
work..look i m not good at expressing..so u can't say it as mumbling..ok!!! ..i treat everybody as equal..and i thought tht asker must have understood wat i m saying...
working i will show u now..it does take my time..
surely adding abt them they r to be positive...
working i will show u now..it does take my time..
surely adding abt them they r to be positive...
3 Mar 2008 13:46:31 IST
Like
0 people liked this
(x1+x2+....xn)/n>=(x1...xn)^1/n=a
also
(x1+k+x2+k+....xn+k)/n=(x1+x2+...xn)/n + k >=[(x1+k)(x2+k)....(xn+k)]^1/n
hence
(x1+x2+...xn)/n >=[(x1+k)(x2+k)....(xn+k)]^1/n - k ------1
also
(x1+x2+...xn)/n >=a -------2
now i don't know what i am doing, ie i don't know what i have proved through the above.
help req!
also
(x1+k+x2+k+....xn+k)/n=(x1+x2+...xn)/n + k >=[(x1+k)(x2+k)....(xn+k)]^1/n
hence
(x1+x2+...xn)/n >=[(x1+k)(x2+k)....(xn+k)]^1/n - k ------1
also
(x1+x2+...xn)/n >=a -------2
now i don't know what i am doing, ie i don't know what i have proved through the above.
help req!
4 Mar 2008 18:52:20 IST
Like
0 people liked this
That is why it is wrong. When you use AM-GM on each product, the minimum occurs when x1 = k; x2 = k, ..;xn=k. But if all xi are equal, they are equal to a
k.
k.That is why the minimum is not 2
(ka)n although it is true that
(ka)n although it is true that(x1+k) (x2+k)...(xn+k) > 2(ka)n
(ka)n The answer you are all getting that the minimum is (a+k)n is the right one. But, I would like to see the justification.
5 Mar 2008 19:23:15 IST
Like
0 people liked this
finally!
What vineet is saying is that you can expand the product
(x1+k)(x2+k)...(xn+k) = kn+kn-1(x1+x2+...+xn)+kn-2(x1x2+x2x3+..+xn-1xn)+...+x1x2x3...xn
Now, AM-GM can be used on each of the summands. It might seem like a tedious affair, but we can cut it short by noticing that the minimum condition occurs when x1=x2=...=xn=a
Hence the minimum is (a+k)n attained when x1=x2=...=xn=a
Now, can you find an upper bound for this product?
Free Sign Up!
Preparing for IIT-JEE ?
Arihant Revision Package for IIT JEE - Books, Practice Tests + Rank Predictor
@ INR 1,995/-
For Quick Info
Find Posts by Topics
Physics.
Topics
9390
20087
3671
2186
7614
2826
2636
Mathematics.
Chemistry.
Biology
Institutes
Parents
Board
Fun Zone
|
|
|
|
|
| 35,229,755 Engineering Questions Attempted |
6,69,530 Visitors Monthly |
79 Engg. Exam Covered |
1916 Students Currently Online |
Connect with Us  |
 |
Vriti Communities
Privacy Policy © Vriti 2011
or simply (a+k)^n...
i think question woulkd have been abt maximum..because minimum we just can express...and not simplify...
because for maximum of products..we have to use h.m-g.m and calculating the sum of 1/ (x1 +k) ..and so on ..is not possible..(generally)..