Question

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Answer 1

Whats your question?

Answer 2

If \[a _{i}>0 \forall i \in N\]
such that \[\prod_{n}^{i=1}a _{i}=1\]
Then prove that \[(1+a _{1})(1+a _{2})(1+a _{3})(1+a _{4}).....(1+a _{n}) \ge 2^{n}\]

Answer 3

@ikram002p @ganeshie8

Answer 4

What does the sign inverted u mean

Answer 5

product of terms

Answer 6

ive seen this before :O

Answer 7

its like \(\sum \),

\[\large \prod_{k=1}^{3}k= (1)(2)(3)\]

Answer 8

\[\large \prod_{k=1}^{3}k^2= (1^2)(2^2)(3^2)\]

Answer 9

etc..

Answer 10

oh

Answer 11

the proof for the original question is simple

Answer 12

Yes then it is easy after i understood what the sign means

Answer 13

i will be back after dinner

Answer 14

yes :) here is an one line proof :


since \(a_i \in \mathbb{N}\), \(a_i+1 \ge 2 \implies \prod \limits_{i=1}^{n}(a _{i}+1)\ge 2^n \)

Answer 15

mmm sure its a proof ?

Answer 16

i thought i had to do this
\(\prod_{n}^{i=1}a _{i}\prod_{n}^{i=1} (1+\dfrac{1}{a_i}) \)

but i got confused for a second sense\( a_i \) is not integer check the product =1

Answer 17

so \(0<a_i\le 1\)
do u agree to this @ganeshie8 ?

Answer 18

wait it can be rational but not integer

Answer 19

forget about the interval

Answer 20

\[(1+a _{1})(1+a _{2})(1+a _{3})(1+a _{4}) \cdots (1+a _{n}) = \prod \limits_{i=1}^{n} (a_i + 1)\]

Answer 21

right ?

Answer 22

yep

Answer 23

ahh i see your point, so we're given that \(a_i \gt 0\),
we're NOT given that its a natural number

Answer 24

yep exactly and sence product of a_i = 1
then for sure it cant be integers
but also it can be <1 or >1

Answer 25

okay then it cannot be an one liner >.<

Answer 26

\(a_i \) should be rationl
even though its confusing
in that case what if \(a_i \le \) 1
then problem solved
but if \(a_i >1 \)like 3/2 or 5/2
then we can do the first line proof u have

but what if \(a_i \) was mixed btw number which >1 or <1 or =1
then we can't conclude anything in this case

Answer 27

lets expand the product

Answer 28

the thing is im too hungry lol xD
so quick i wanna sleep hehe
unless there is a given condition for \(a_i\)

Answer 29

I got it how to do

Answer 30

see ganesh
for example
(1+1/2) (1+1/3) (1+1/4) <2^3
(1+1/3776568)(1+3/2)(1+1/2)(1/45456) <2^4 ( not sure :P)

Answer 31

only this case work , if \(1\le a_i \)
(1+3/2) (1+5/4) (1+7/6) >=2^3

Answer 32

nope we cant , we only can conclude that \(a_i\) is rational

Answer 33

for example
\(\dfrac{1}{2}×\dfrac{2}{3}×\dfrac{3}{1}×\dfrac{1}{1}=1\)

Answer 34

ok

Answer 35

gtg nw ,
have a nice day

Answer 36

\[(1+a _{1})(1+a _{2})(1+a _{3}).....(1+a _{n})\ge2^{n}[(a _{1}a _{2}a _{3}..a _{n}]^{1/2}\]


so proved