Smallest Possible Value for Inequality: Am I on the Right Track?

[Fellowroot]

Homework Statement

Homework Equations

n/a

The Attempt at a Solution

Here is one attempt

But I'm stuck on this inequality. I can't go further.

and here is another, but I don't know if I proved anything here.

Really looking if anyone could help me on this or if I'm on the right track. Thanks.

 

[gopher_p]

There are two components to the problem; (1) find a candidate for the solution and (2) prove that candidate is the right one.

For (1) I recommend throwing away all this fancy schmancy algebra and calculus and just roll up your sleeves and try a few things. Also, recognizing that your problem is equivalent to maximizing $\frac{1}{n}+\frac{1}{m}+\frac{1}{k}$ subject to $n,m,k$ distinct and $\frac{1}{n}+\frac{1}{m}+\frac{1}{k}<1$ might make some of this work a little more manageable.

Once you've found a triplet that works, try to prove that it's the best triplet. Don't get fancy, just think about it. If need be, find other triplets that work (in the sense that $n,m,k$ are distinct and $\frac{1}{n}+\frac{1}{m}+\frac{1}{k}<1$) and try to see why your triplet is better.

 

[Ray Vickson]

Homework Statement

Homework Equations

n/a

The Attempt at a Solution

Here is one attempt

But I'm stuck on this inequality. I can't go further.

and here is another, but I don't know if I proved anything here.

Really looking if anyone could help me on this or if I'm on the right track. Thanks.

No, you are on the wrong track: you cannot take derivatives with respect to discrete (integer-valued) variables like n, m and k. Derivatives need continuous variables, and you don't have those in this problem.

 

[HallsofIvy]

And even if m, n, and k were continuous variables, you certainly cannot take the derivative with respect to three different variables as you did here.

 

[Fellowroot]

Thanks for pointing that out because I just remembered that it would be an implicit differentiation if I did take the derivative and I did not do that.

recognizing that your problem is equivalent to maximizing 1/n+1/m+1/k

Thanks for the hint, but I don't quite understand it. I thought I was trying to minimize it not maximize it.

 

[dirk_mec1]

In this case if you maximize 1/n+1/m+1/k (while still being smaller than 1) you minimize your goal function, right?

 

[Fellowroot]

Thank you dirkmec1 I now understand!