Mean and weighted mean comparison

  • Thread starter gianeshwar
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  • #1
gianeshwar
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Dear Friends!
Is weighted mean greater than mean?
Actually i need to prove (c-b)f(a)+(b-a)f(c)>(c-a)f(b) when when f is continuous and strictly increasing in [a,c],where a<b<c.
 

Answers and Replies

  • #2
jbriggs444
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Is what you are trying to prove even true? Can you check against some simple example functions?
 
  • #3
gianeshwar
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Dear friend!Thank You for response!
I have checked it and it comes true for f(x) equal to x square for the positive real domains.
 
  • #4
jbriggs444
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Have you checked against the identify function, f(x) = x?

And have you compared the formula in question to linear interpolation?
 
  • #5
Stephen Tashi
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I have checked it and it comes true for f(x) equal to x square for the positive real domains.

Try [itex] f(x) = \sqrt{x},\ a = 0,\ b = 1,\ c = 9 [/itex]
 
  • #6
gianeshwar
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Thank You dear friends!
I had realized my mistake and an modifying the question.
Now ingnore first part of question and assume that if in addition it is specified that f'(x) is strictly increasing then the cnjecture seems true.BUT I NEED GENERAL PROOF.
 
  • #7
jbriggs444
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The mean value theorem should have something to contribute. Consider the slopes of the lines from (a,f(a)) to (b,f(b)), from (b,f(b)) to (c,f(c)) and from (a,f(a)) to (c,f(c)).

If the inequality fails to hold then you should be able to demonstrate the existence of two points x and y such that x > y and f'(x) <= f'(y)
 
  • #8
gianeshwar
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Thank You all friends!
Ultimately could solve as follows:
f(b)-f(a)/b-a less than f(c)-f(b)/c-b,because f'(p) less than f'(q) due to f'(x) strictly increasing and p in (a,b) and q in (b,c).
 

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