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If a1.a2.a3...an=1 and n>=3,

the prove that: (1+a1)1(1+a2)2....(1+an)n>nn

Just to add, I thought about taking two assumptions of n=even and n=odd but got no where! :(

I could be wrong but I think there needs to be more information given. we're given that the product of a1 through an is 1 but not given info on the values of ai. Based off the info given if I let a1 = -2, a2 = 3, a3 = -1/3, and a4 = 1/2 then a1a2a3a4 = 1 but (1+a1)1(1+a2)2(1+a3)3(1+a4)4 < 44 . I think if were given that ai > 0 for all i then I think we could prove this.

Please let me know if there is anymore information then I could come up with a proof. It looks like a typical Proof by Mathematical Induction type problem

All a i - s have to be positive numbers.

All a i - s have to be positive numbers.

### 2 Answers by Expert Tutors

Grigori S. | Certified Physics and Math Teacher G.S.Certified Physics and Math Teacher G.S.
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For all a i > 0 we can rewrite the inequality in the following way:

∏  i=1,n  (1 + ai) i ≥ nn

(∏  symbol of a product of n consequitive terms).

Using the symbol of factorial we can rewrite this inequality again

(1+a1) .....(1+an) ≥ nn/ n!

According to the theorem about arithmetic and geometric means

(1 + ai)/2 ≥ √ai

we have

(1+a1)....(1+an) ≥ 2n sqrt(a1a2....an) = 2n

If 2n (n!) ≥ nn, then it proves the statement. But this inequality is true for n ≤ 5.

The best way to prove the staement is to use the principle of mathematical induction.

The starement is true for n=1 (it is evidently). Then assume that the statement is true for any "n", and check if this is true for "n+1". For "n+1" we have

(1 + a1) ......(1+an)(1+a n+1) (n+1)! ≥ (n+1)n+1

We can transform the left side of the inequality using the statement for "n":

(1+a1) .....(1+an)(1+a n+1) (n+1)! ≥ nn (1+a n+1) (n+1)  ≥   (n+1)n (n+1)

The right side can be written in the form

(n+1)n (n+1) = nn [1+(1/n)]n (n+1)

If we look at last two inequalities we can see that the original inequality can be proven if we prove that

1 + a n+1 ≥ [1 + (1/n)] n

or

(1 + a n+1)1/n ≥ 1 + (1/n)

Now take into account that for "n+1"

a1a2......an a n+1 = 1 (according to the original condition)

That means

a n+1 = 1/(a1 ....an)

and for large n we can write (approximately)

(1 + a n+1) 1/n ≥ 1 + 1/n (a1 ...an)

If we assume that a1 ...an ≤ 1 then

1 + 1/[n (a1 ...an)] ≥ 1 + (1/n)

and the inequality is proven. But we need more information about {ai} to be more specific in our  conclusions.

Robert J. | Certified High School AP Calculus and Physics TeacherCertified High School AP Calculus and Ph...
4.6 4.6 (13 lesson ratings) (13)
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It doesn't work. Counter example: a1 = -1, a2 = -1, and a3 = 1.

Do you have more restrictions?
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