Pls. Help.......Pls. Help...

so thats the mistake. well there is some other method and i suggest u figure it out!

We 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

LHS >0, a,b,c R fractions!!!

@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

sahi hai....sry i was thinking wrong

But wat U did isnt always true.

It can be either way.

Coz Math is fact not chance.

One exception proves the statement wrong[5]

here
abc is a fraction and so is (1-a)(1-b)(1-c) . So I dont see any mistake in dividing !!!

C'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.......

well, its simple am gm use na

[ (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

well is it a+b+c=2 or a+b+c≥2 ? once check..

dats absolutely fine.....but pata toh chale wheres the mistake so that kahin aur wahi galti na ho.

dude 5 days b4 JEE is not da time for PROOFS,,,,,,,,,,,

RELY ON SORTCUTS

.75,.75,.5

it cums >8 obviously

rk take an eg na.....

pls. solve and post soln.

arre usi mein toh doubt hai.
I'm getting LHS ≤ 8 instead of ≥ 8 . I'm not able to find my mistake.