You need 3 things for maths.

Algebra Level 5

Let $n$ be a non-zero real number (not necessarily positive). Find the minimum value of

$\prod_{i = 1}^{8}(n^{2^{i}} + 1) + \prod_{i = 1}^{8}(\frac{1}{n^{2^{i}}} + 1).$

According to my thoughts, you need 3 things for solving math problems:

Knowledge
Imagination
Solve this problem to figure that one out.

Hint: The third word starts with $T$ and ends with $S$ . It may depend on each person whether this hint has actually helped or not.

The answer is 512.

1 solution

Sanchayapol Lewgasamsarn
May 22, 2014

ฺBy noticing: $(n^{2} + 1)(n^{4} + 1) = n^{6} + n^{4} + n^{2} + 1$

$(n^{2} + 1)(n^{4} + 1)(n^{8} + 1) = n^{14} + n^{12} + n^{10} + n^{8} + n^{6} + n^{4} + n^{2} + 1$

And so, in a similar manner,

$\prod_{i = 1}^{8}(n^{2^{i}} + 1) = n^{510} + n^{508}+ n^{506} + ... + 1$ (206 Terms)

And in a similar manner,

$\prod_{i = 1}^{8}(\frac{1}{n^{2^{i}}} + 1) = \frac{1}{n^{510}} + \frac{1}{n^{508}} + ... + 1$ (206 Terms)

By the A.M.-G.M. Inequality

$\prod_{i = 1}^{8}(n^{2^{i}} + 1) + \prod_{i = 1}^{8}(\frac{1}{n^{2^{i}}} + 1) \geq 512\cdot 1$

$\prod_{i = 1}^{8}(n^{2^{i}} + 1) + \prod_{i = 1}^{8}(\frac{1}{n^{2^{i}}} + 1) \geq \boxed{512}$

Note that we can use the A.M.- G.M. Inequality because all members used in the Inequality are positive because $n \neq 0$ and $n^{2i} \geq 0$ when $i$ is a positive integer.

And, the word is TOUGHNESS!

Nice observation.

You can approach this with AM-GM directly:

$\prod_{i=1}^8 ( n ^ {2^i} + 1 ) + \prod_{i=1}^8 ( \frac{1}{n ^ {2^i} } + 1 ) \geq 2 \sqrt{ \prod_{i=1}^8 ( n ^ {2^i} + 1 ) ( \frac{1}{n ^ {2^i} } + 1 )} \geq 2 \sqrt{ \prod_{i=1}^8 4} = 512.$

Calvin Lin Staff - 7 years ago

Could you please tell me how you simplified the geometric mean to get 512?

Saksham Bansal - 6 years, 10 months ago

Notice how sir Lin did not write an equality sign; the simplification of the geometric mean simply goes as follows: $\prod\limits_{ i=1 }^{ 8 }{ { (n }^{ { 2 }^{ i } }+1){ (\frac { 1 }{ { n }^{ { 2 }^{ i } } } }+1) } =\prod\limits_{ i=1 }^{ 8 }{ { (\frac { { n }^{ { 2 }^{ i } } }{ { n }^{ { 2 }^{ i } } } }+{ n }^{ { 2 }^{ i } }+\frac { 1 }{ { n }^{ { 2 }^{ i } } } +1) }$ $=\prod\limits_{ i=1 }^{ 8 }{ (2+{ n }^{ { 2 }^{ i } }+\frac { 1 }{ { n }^{ { 2 }^{ i } } } ) }$ I think we must simply assume that this will have its minimum when $n = 1$ , resulting in Lin's expression after the geometric mean. Maybe multiplying ${n}^{ {2}^{i} } + \frac { 1 } { {n}^{ {2}^{i} } }$ with ${ n }^{ { 2 }^{ i } }$ shows the minimum value, as ${ n }^{ { 2 }^{ i } }$ is always positive, to occur at said $n = 1$ . Q.E.D? :P

Ralph Schraven - 6 years, 10 months ago

Out of curiosity, what does the word have to do with problem?

Calvin Lin Staff - 7 years ago

It should have taken a really long time to expand the 8 polynomials' product. Hahaha

Sanchayapol Lewgasamsarn - 7 years ago

How did you get 512?

Saksham Bansal - 6 years, 10 months ago

The word could be "twos" as well. Setting $n^{2^i} + 1 = 2$ works as well.

Josh Speckman - 6 years, 10 months ago

Where does the 512 come from, and how is that the gm if there is no n root?

Trevor Arashiro - 6 years, 10 months ago

The claim that there are 206 terms in the expression next to that claim is not correct. There are $\frac{510}{2}+1 = 256$ terms in both of the expressions where the claim is made.

Ralph Schraven - 6 years, 10 months ago

This problem can quite easily be solved by simply assuming that the two products have their minimum when n is set to 0. Proving this would be more difficult, in which case AM-GM really shows to be impeccable in being both productive and analytically correct, unlike guessing the right answer.

Ralph Schraven - 6 years, 10 months ago

small typing error .it is 256 in place of 206 .well solved.

Adarsh Kumar - 6 years, 9 months ago

You need 3 things for maths.

The answer is 512.

1 solution

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