## anonymous 2 years ago .

1. linda3

2. anonymous

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

3. anonymous

@ikram002p @ganeshie8

4. anonymous

5. ganeshie8

product of terms

6. ikram002p

ive seen this before :O

7. ganeshie8

its like $$\sum$$, $\large \prod_{k=1}^{3}k= (1)(2)(3)$

8. ganeshie8

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

9. ganeshie8

etc..

10. anonymous

oh

11. ganeshie8

the proof for the original question is simple

12. anonymous

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

13. anonymous

i will be back after dinner

14. ganeshie8

yes :) here is an one line proof : since $$a_i \in \mathbb{N}$$, $$a_i+1 \ge 2 \implies \prod \limits_{i=1}^{n}(a _{i}+1)\ge 2^n$$

15. ikram002p

mmm sure its a proof ?

16. ikram002p

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

17. ikram002p

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

18. ikram002p

wait it can be rational but not integer

19. ikram002p

20. ganeshie8

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

21. ganeshie8

right ?

22. ikram002p

yep

23. ganeshie8

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

24. ikram002p

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

25. ganeshie8

okay then it cannot be an one liner >.<

26. ikram002p

$$a_i$$ 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

27. ganeshie8

lets expand the product

28. ikram002p

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

29. anonymous

I got it how to do

30. ikram002p

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)

31. ikram002p

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

32. ikram002p

nope we cant , we only can conclude that $$a_i$$ is rational

33. ikram002p

for example $$\dfrac{1}{2}×\dfrac{2}{3}×\dfrac{3}{1}×\dfrac{1}{1}=1$$

34. ganeshie8

ok

35. ikram002p

gtg nw , have a nice day

36. anonymous

|dw:1404314484947:dw| $(1+a _{1})(1+a _{2})(1+a _{3}).....(1+a _{n})\ge2^{n}[(a _{1}a _{2}a _{3}..a _{n}]^{1/2}$ |dw:1404314863576:dw| so proved

37. ganeshie8

nice :) AM/GM inequality to the rescue !!

38. anonymous

yes, just didn't knew what that sign meant

39. ikram002p

cool O.O never seen this before

40. ganeshie8

AM-GM inequality : $$\dfrac{a+b}{2} \ge \sqrt{ab}$$

41. ganeshie8

@ikram002p

42. ikram002p

ic .. i dont think that ive seen it before :D