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- Feb 14, 2012

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Determine all possible values of $f$.

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- #1

- Feb 14, 2012

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Determine all possible values of $f$.

- Jan 30, 2018

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- Feb 14, 2012

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- Mar 31, 2013

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$\sum_{1=1}^8 x_i = 4\dots(1)$

$\sum_{1=1}^7 \sum_{j=i+1}^8 x_ix_j = 7\cdots(2)$

$\prod_{1=1}^8 x_i = f\cdots(3)$

From (1) and (3) using AM GM inequality as all are positive taking AM ad GM of the elements of parameters of equation (1)

$AM = \frac{\sum_{1=1}^8 x_i}{8} = \frac{4}{8} = \frac{1}{2}\dots(4)$

$GM= \sqrt[8]{\prod_{1=1}^8x_i}=\sqrt[8] f\cdots(5)$

As AM is larger than equal to GM we get from (4) and (5)

$f<= \frac{1}{2^8}\cdots(6)$

From (2) and (3) using AM GM inequality taking AM ad GM of the elements of parameters of equation (2)

$AM = \frac{\sum_{1=1}^7 \sum_{j=i+1}^8 x_ix_j }{28}= \frac{7}{28} = \frac{1}{4}\cdots(7)$ as there are 28 terms

$GM = \sqrt[28]{\prod_{1=1}^7 \prod_{j=i+1}^8 x_ix_j }= \sqrt[28]{(\prod_{1=1}^8)^7}= \sqrt[28]{f^7}= \sqrt[4]{f} \cdots(8)$

As AM is larger than equal to GM we get from (7) and (8)

$f<= \frac{1}{4^4})$

or $f<= \frac{1}{2^8}\cdots(9)$

From (6) and (9) we have $f<= \frac{1}{2^8}$

Actually f becomes $\frac{1}{2^8}$when all the $x_i$ are $\frac{1}{2}$

As any $x_i$ can be as low as possible but not zero we we get the condition

$0 < f<= \frac{1}{2^8}$

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- Feb 14, 2012

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- Mar 31, 2013

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2nd term 7. not 28

- Mar 31, 2013

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$\sum_{i=1}^8 x_i = 4\dots(1)$

$\sum_{i=1}^7 \sum_{j=i+1}^8 x_ix_j = 7\cdots(2)$

$\prod_{i=1}^8 x_i = f\cdots(3)$

We have

$\sum_{i=1}^8 x_i^2= (\sum_{i=1}^8 x_i)^2 - 2 \sum_{i=1}^7 \sum_{j=i+1}^8 x_ix_j $

$= 4^2 - 2 * 7 = 2$

or $\sum_{i=1}^8 x_i^2 = 2$

Subtracting (1) from above

$\sum_{i=1}^8 (x_i^2 - x_i) = -2$

adding $\frac{1}{4}$ to each term on LHS that is 2 and adding 2 on RHS we get

$\sum_{i=1}^8 (x_i^2 - x_i + \frac{1}{4}) = 0$

or $\sum_{i=1}^8 (x_i - \frac{1}{2})^2 = 0$

so $x_i = \frac{1}{2}$ for each i.

this satisfies the criteria that $x_i$ is positive

putting this in (3) we get $f = \frac{1}{\sqrt[8]2}$