IIT JEE , AIEEE Forum - Mathematics , Algebra

vineetnegi

(a/(1-a))*(b/(1-b))*(c/(1-c))

when

a+b+c=2

min?

konichiwa2x

You should be able to take it from here, no doubt?

If you are not able to, nudge me.

edit: Here is the complete solution.

if a,b,c restricted to positive values less than 1 then

a=1-x, b = 1-y, c=1-z ;0 < x,y,z < 1 give

where x+y+z=1

hence,

and as each factor is of the form

Equality holds when x=y=z = 1/3 when a=2/3, b=2/3, and c=2/3

vineetnegi

go on

akhil_o

for (1-a),(1-b),(1-c) AM is 1/3
so using AM>=GM
(1-a)(1-b)(1-c)<=(1/3)^3=1/27

a,b,c AM =2/3
so
abc<=(2/3)^3
=8/27

eqn= abc/(1-a)(1-b)(1-c)
=8/27/1/27
=8
so min vaue

akhil_o

but if 0 is allowed
min value is 0

computer001

8,when a=b=c=2/3

hsbhatt

Easier method:

$a+b+c =2 \\ \\ \text {or} \ a = (1-b) + (1-c) \\ \\ \text {Hence by AM-GM Inequality} \ a \geq 2 \sqrt{(1-b)(1-c)} \\ \\ \text {Similarly} \ b \geq 2 \sqrt{(1-c)(1-a)} \\ \\ \text {and} \ c \geq 2 \sqrt{(1-a)(1-b)} \\ \\ \text {Multiplying the three relations} \ abc \geq 8 (1-a)(1-b)(1-c) \\ \\ \text {Hence the minimum value of} \frac {abc} {(1-a)(1-b)(1-c)} \ \text{is 8}$

konichiwa2x

By the way there is no minimum unless you specify .

Most of you would say this is not so important, but it is!

Consider this case :

Let

.

Then

.

As

, then

decreases without bound, and

. So, let

and

as

and the expression decreases without bound.
_________________

hsbhatt

u r right, but u cud just proceed with the assumption that all qts involved r +ve.

konichiwa2x

well, I am never happy unless things are specified clearly. Often I end up spending too much time on problems, due to lack of clarity in the question. Even for this problem, I discussed it for all possible cases. I guess its my weakness..

sandeepramesh

can someone post a new question?

konichiwa2x

sure, what topic?

sandeepramesh

anything except co geo and 3 D geo

konichiwa2x

alright, here is a nice integral.

http://www.goiit.com/posts/list/0/integration-a-nice-integral-47108.htm

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