- #1

- 93

- 0

Given that a, b, c, d are positive integers and a+b=c+d.

Prove that if a∗b < c∗d,

then a∗log(a)+b∗log(b) > c∗log(c)+d∗log(d)

How do I do it?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter japplepie
- Start date

- #1

- 93

- 0

Given that a, b, c, d are positive integers and a+b=c+d.

Prove that if a∗b < c∗d,

then a∗log(a)+b∗log(b) > c∗log(c)+d∗log(d)

How do I do it?

- #2

RUber

Homework Helper

- 1,687

- 344

So if AB < CD, then C and D are more central...which gives you A < C ≤ D < B.

From there use the fact that the log function is concave down.

Please post a little bit more about what you have tried, and where you are stuck.

- #3

- 93

- 0

How do I use the concave down point?

- #4

- 93

- 0

Given a new constrant that A+B = C+D = 1

Does showing that:

d[ -1(a*log(a)+(1-a)*log(1-a)) ] / d[a] * d[ a*(1-a) ] / d[a] to be always greater than or equal to zero prove the original claim?

Since satisfying this means that the two functions grow and shrink together (albeit not in the exact amount).

y=-1/log(2)*(x*log(x)+(1-x)*log(1-x)) {[0,1]} // the a*log(a)+b*log(b)

y=x*(1-x) {[0,1]} // the a*b

Does showing that:

d[ -1(a*log(a)+(1-a)*log(1-a)) ] / d[a] * d[ a*(1-a) ] / d[a] to be always greater than or equal to zero prove the original claim?

Since satisfying this means that the two functions grow and shrink together (albeit not in the exact amount).

y=-1/log(2)*(x*log(x)+(1-x)*log(1-x)) {[0,1]} // the a*log(a)+b*log(b)

y=x*(1-x) {[0,1]} // the a*b

Share: