| Posted on Wednesday, 30 July, 2003 - 11:56 am: |
Could someone please explain how this inequality works and show me an example of how to use it?
| Posted on Wednesday, 30 July, 2003 - 12:03 pm: |
The particular problem I'm stuck with is:
Prove that
where
| Posted on Wednesday, 30 July, 2003 - 12:26 pm: |
There is a terse description here .
The easiest example to visualise is the case when there are 2 terms: The idea is that if you have a convex function [e.g. y=x2 ] then if I pick any two points [say (-1,1) and (3,9)] and draw a straight line between them, then if I pick the mid x point [i.e. (-1+3)/2=1], then the y value as read off the convex function [12 =1]is less than the corresponding point read off from the straight line [5]. The inequality is reversed if the function is concave.
If your case, we observe that x.ln(x) is real and continuous for x> 0. Further (looking at the second derivative) we see that the function is convex. Therefore the 3 term inequality looks like
1/3(f[p] + f[q] + f[r]) > = f[(p+q+r)/3]
Substitute p=a, q=b, r=c and f[x] = x.ln[x] we have
1/3(f[a] + f[b] + f[c]) > = f[(a+b+c)/3]
1/3(a.ln[a] + b.ln[b] + c.ln[b]) > = (a+b+c)/3.ln[(a+b+c)/3]
Which is your result.
Andre
| Posted on Wednesday, 30 July, 2003 - 03:30 pm: |
Ah yes
I wasn't getting how to substitute the function but now it is clear
| Posted on Wednesday, 30 July, 2003 - 11:51 pm: |
The link says that the AM-GM inequality is a special case of Jensen's inequality. How does this work?
| Posted on Thursday, 31 July, 2003 - 12:21 am: |
Jensen's inequality says that if a function is concave (2nd derivative is negative) that the average of the function is less than the function of the average.
So, since ln is a concave function, (ln(a1 ) + ln(a2 ) + ... + ln(an ))/n < = ln((a1 +a2 +...+an )/n)
Take the exponential function of both sides to see that the geometric mean is less than or equal to the arithmetic mean.
| Posted on Thursday, 31 July, 2003 - 10:54 pm: |
Thank you.