# Mean and weighted mean comparison

1. Dec 9, 2014

### gianeshwar

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.

2. Dec 9, 2014

### jbriggs444

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

3. Dec 9, 2014

### gianeshwar

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. Dec 9, 2014

### jbriggs444

Have you checked against the identify function, f(x) = x?

And have you compared the formula in question to linear interpolation?

5. Dec 9, 2014

### Stephen Tashi

Try $f(x) = \sqrt{x},\ a = 0,\ b = 1,\ c = 9$

6. Dec 9, 2014

### gianeshwar

Thank You dear friends!
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. Dec 9, 2014

### jbriggs444

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. Dec 11, 2014

### gianeshwar

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).