39so thats the mistake. well there is some other method and i suggest u figure it out!
a,b,c are 3 nos. such that 1 > a,b,c > 0 and a + b + c = 2
Then show that \frac{a}{(1-a)}\frac{b}{(1-b)}\frac{c}{(1-c)} ≥ 8
This may be very simple but I'm getting LHS ≤ 8 instead of ≥ 8 . I'm not able to find my mistake. So pls. help.
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24 Answers
39
1We R to find the minimum value of the given a/1-a~~ thing
This would happen when each of the term is minimum.
ie., a=b=c=2/3
Substituting this in the given exp gives 8.
So the value is always ≥8
11@b555.........
kahaan se lhs < 0 ?????????
lhs = {(1-a)+(1-b)+(1-c)}/3 ]3 = {3-(a+b+c)}/3 ]3 = {3-2}/3 ]3 = (1/3)3 = 1/27
1But wat U did isnt always true.
It can be either way.
Coz Math is fact not chance.
One exception proves the statement wrong[5]
11here
abc is a fraction and so is (1-a)(1-b)(1-c) . So I dont see any mistake in dividing !!!
11@b555 & kalyan
What about this........
1/4 < 1/2
1/6 < 1/4
--------------
(1/4)/(1/6) = 3/2
(1/2)/(1/4) = 2
3/2 < 2
1C'mon Dude how can U do that[7][7]
See ,
1<2
2<6
Now does that mean
1/2 <1/3 [7]
Think again.......
11[ (a + b + c)/3 ]3 ≥ abc
abc ≤ 8/27 ...................(1)
[ {(1-a)+(1-b)+(1-c)}/3 ]3 ≥ (1-a)(1-b)(1-c)
(1-a)(1-b)(1-c) ≤ 1/27 ....................(2)
Dividing (1) / (2) Shayad here's the mistake [12][12]
abc/(1-a)(1-b)(1-c) ≤ 8
11dats absolutely fine.....but pata toh chale wheres the mistake so that kahin aur wahi galti na ho.
21dude 5 days b4 JEE is not da time for PROOFS,,,,,,,,,,,
RELY ON SORTCUTS
11ya i took 2/5,4/5 & 4/5 . usse i got the rite ans. but taking a,b,c i think i'm making some mistake.pata nahi kahaan.
pls. try then i'll say wat i did.
11arre usi mein toh doubt hai.
I'm getting LHS ≤ 8 instead of ≥ 8 . I'm not able to find my mistake.