I See Some Relation
If the minimum value of the expression S = x 1 0 0 0 + x 9 0 0 + x 9 0 + x 6 + x 1 9 9 6
is β for positive real x , find 1 0 0 0 β + 6 .
The answer is 8.
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1 solution
You have only shown that lower bound is 2000, you still need to show that the minimum value of 2000 is possible.
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Umm... I don't know how else to show that it is the minimum value, but will this work?
Observe the behaviour of the function at x = 1 . We can observe that when we go towards values less that 1, the value of 1 9 9 6 / x > 1 9 9 6 and the value of the function is greater that 2000. Now for values greater than 1, the value of the function is still greater than 2000. So x = 1 must be the minima. (Actually I substituted the values)
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Hint: If AM(a,b, c) >= GM(a,b,c)
then when does the equality holds?
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@Pi Han Goh – Ok, do i havs to show the equality case?
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@Sravanth C. – Yes, exactly that.
S = x 1 0 0 0 + x 9 0 0 + x 9 0 + x 6 + x 1 9 9 6
lo g S = lo g x ( 1 9 9 6 − 1 9 9 6 )
lo g S = 0
S = 1
:P
Applying AM-GM on < x 1 0 0 0 , x 9 0 0 , x 9 0 , x 6 , 1 / x , ⋯ , 1 / x > we arrive at 2 0 0 0 S ≥ 1 1 / 5 S ≥ 2 0 0 0
If x 1 0 0 0 = x 9 0 0 = x 9 0 = x 6 = 1 / x = ⋯ = 1 / x then x = 1 and S = 2 0 0 0 which satisfies the above inequality.
And 1 0 0 0 β + 6 = 1 0 0 0 2 0 0 0 + 6 = 8 .