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 }
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?
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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.
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