Hard Math Question - stuck, please help.

Question

can you please help me prove this:

 

If a₁.a₂.a₃...a_n=1 and n>=3,

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

Answer 1

For all a _i > 0 we can rewrite the inequality in the following way:

                                                 ∏  _i=1,n  (1 + a_i) i ≥ nⁿ   

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

Using the symbol of factorial we can rewrite this inequality again

                                                     (1+a₁) .....(1+a_n) ≥ nⁿ/ n!

According to the theorem about arithmetic and geometric means

                                                   (1 + a_i)/2 ≥ √a_i

we have

                                   (1+a₁)....(1+a_n) ≥ 2ⁿ sqrt(a₁a₂....a_n) = 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+a_n)(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+a_n)(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......a_n a _n+1 = 1 (according to the original condition)

That means

                                            a _n+1 = 1/(a₁ ....a_n)

and for large n we can write (approximately)

                                 (1 + a _n+1) ^1/n ≥ 1 + 1/n (a₁ ...a_n)

If we assume that a₁ ...a_n ≤ 1 then

                               1 + 1/[n (a₁ ...a_n)] ≥ 1 + (1/n)

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

 

Answer 2

It doesn't work. Counter example: a₁ = -1, a₂ = -1, and a₃ = 1.

Do you have more restrictions?