# help for proofs (1 Viewer)

#### fan96

##### 617 pages
Where did you get this question?

You probably wouldn't be able to get this result from just as

is not actually true, even taking . You'd have to at least start a bit further back.

I doubt this is the intended answer but one hence-y way could be to cook up another integral in a similar fashion that gives you the inequality you want, e.g.

#### CM_Tutor

##### Well-Known Member
I'm pretty sure that the required substitution is or in which case the result becomes . We then need to bring in the integral result or some modification (I tried trapezoidal and it doesn't get us there) to get the final result.

I agree with mathbrain, it is certainly a true result as has for all and so (and thus ) for all .

• DrDawn and mathsbrain

#### mathsbrain

##### Member
I'm pretty sure that the required substitution is or in which case the result becomes . We then need to bring in the integral result or some modification (I tried trapezoidal and it doesn't get us there) to get the final result.

I agree with mathbrain, it is certainly a true result as has for all and so (and thus ) for all .
damn...this is tough, can anyone help?

#### CM_Tutor

##### Well-Known Member
Extending from my previous post and correcting an error...

And use the substitution :

Now, returning to the starting inequality, we had:

Applying the substitution :

Now, equations (1) and (2A) can be combined to conclude that ... which it true, but not terribly useful:

The approach can be modified, however, increasing or decreasing one side of an inequality where the change cannot falsify the statement. This is akin to saying that if and that then I can say with certainty that . In this case, I start by noting that allows me to alter {2B):

Other variants include modifying (1A)'s term like :

These are getting closer, though the negative coefficient on the is a problem. As an aside, we would be there if we can get as this could be integrated from 0 to to give , which is the required result.

I don't see how this approach is going to yield any result where has positive coefficients on any terms in . I've also been thinking about the geometric interpretations here:
• Consider the function plotted from to where . Add divisions (for upper and lower rectangles) with width of 1 unit.
• is the area of the resulting lower rectangles, are the upper rectangles, and so the inequality
• Taking we have a rectangle with one side on the -axis covering and one side on the -axis covering
• The integral exists entirely within the rectangle, so that the area of the rectangle () is divided into the part under the curve (with area ) meaning the area above the curve (which is also the area under the curve against the -axis has area
• This provides insight into some of the inequalities and the reason for the substitution, but does not yield a solution.
• I do wonder, though, if we need to work against the -axis and the diagonal for the large rectangle, and seek that the area against the -axis to be greater than - but while still using the original result to satisfy the "hence" requirement.