Problem 18

Algebra Level 4

$\begin{aligned} A &= &(1+a_2)^2 (1+a_3)^3 \cdots (1+a_{2016})^{2016} \\ B &=& 2016^{2016} \end{aligned}$

Given that $a_2, a_3, \ldots, a_{2016}$ are non-negative real numbers such that their product is 1, and let $A$ and $B$ as described above.

Which of the following equation/inequation must be true?

This problem inspired of IMO problem in a few year back.

This problem is in this Set .

A = B

A>B

There is insufficient information

A<B

1 solution

Son Nguyen
Feb 26, 2016

We will prove:

$(1+a_{2})^2(1+a_{3})^3...(1+a_{n})^n> n^n$

By applying AM-GM inequality we have: $(a_{k}+1)^k=(a^k+\underset{k-1 times}{\frac{1}{k-1}+\frac{1}{k-1}+...+\frac{1}{k-1}})^k\geq k^k\times a_{k}\times \frac{1}{(k-1)^{k-1}},k=\overline{2,n}$

The equality holds when $a_{k}=\frac{1}{(k-1)!},k=\overline{2,n}$ ,this can't be happen.

Mutiply each side $(n-1)$ we get: $(1+a_{2})^2(1+a_{3})^3...(1+a_{n})^n> \frac{2^2}{1^2}\times \frac{3^2}{2^2}...\frac{n^n}{(n-1)^{n-1}}\times a_{2}a_{3}...a_{n}=n^n$

So the answer must be $>$ with $n=2016$ .

This problem inspired of 53 third IMO 2012 Mar Del Plata,Argentina,South America.

Great solution. There's a typo in the second last line though, it should be $3^3$ , not $3^2$ .

Harsh Poonia - 1 year, 11 months ago

Problem 18

This problem inspired of IMO problem in a few year back.

This problem is in this Set .

1 solution

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