- #1

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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.

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- Thread starter gianeshwar
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- #1

- 222

- 12

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.

- #2

jbriggs444

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Is what you are trying to prove even true? Can you check against some simple example functions?

- #3

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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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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

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I had realised 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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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

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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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