can you please help me prove this:

If a_{1}.a_{2}.a_{3}...a_{n}=1 and n>=3,

the prove that: (1+a_{1})1(1+a_{2})2....(1+a_{n})n>n^{n
}

can you please help me prove this:

If a_{1}.a_{2}.a_{3}...a_{n}=1 and n>=3,

the prove that: (1+a_{1})1(1+a_{2})2....(1+a_{n})n>n^{n
}

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For all a_{ i }> 0 we can rewrite the inequality in the following way:

∏ _{i=1,n} (1 + a_{i}) i ≥ n^{n}

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

Using the symbol of factorial we can rewrite this inequality again

(1+a_{1}) .....(1+a_{n}) ≥ n^{n}/ n!

According to the theorem about arithmetic and geometric means

(1 + a_{i})/2 ≥ √a_{i}

we have

(1+a_{1})....(1+a_{n}) ≥ 2^{n
}sqrt(a_{1}a_{2}....a_{n}) = 2^{n}

If 2^{n} (n!) ≥ n^{n}, 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 + a_{1}) ......(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+a_{1}) .....(1+a_{n})(1+a _{n+1}) (n+1)! ≥ n^{n} (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) = n^{n} [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"

a_{1}a_{2}......a_{n} a _{n+1} = 1 (according to the original condition)

That means

a _{n+1 }= 1/(a_{1} ....a_{n})

and for large n we can write (approximately)

(1 + a _{n+1}) ^{1/n} ≥ 1 + 1/n (a_{1} ...a_{n})

If we assume that a_{1} ...a_{n} ≤ 1 then

1 + 1/[n (a_{1} ...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.

It doesn't work. Counter example: a_{1} = -1, a_{2} = -1, and a_{3} = 1.

Do you have more restrictions?

## Comments

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 a

_{1}through a_{n}is 1 but not given info on the values of a_{i}. Based off the info given if I let a_{1}= -2, a_{2 }= 3, a_{3}= -1/3, and a_{4}= 1/2 then a_{1}a_{2}a_{3}a_{4}= 1 but (1+a_{1})1(_{}1+a_{2})2(1+a_{3})3(1+a_{4})4 < 4^{4 }. I think if were given that a_{i}> 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.