# Larsen problem

1. Dec 23, 2007

### ehrenfest

1. The problem statement, all variables and given/known data
Is what I wrote on the left hand margin a counterexample to 1.8.5 part a) ?

EDIT: I meant part b)

2. Relevant equations

3. The attempt at a solution

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Last edited: Dec 24, 2007
2. Dec 24, 2007

### Shooting Star

Yes. (For b).

3. Dec 24, 2007

### ehrenfest

And changing a+b=1 to a+b=2 makes the inequality correct, right?

Also, in part b) of 1.8.7, are they asking me to show that 1/(a+c) < 1/(b+c) + 1/(a+b) ?

4. Dec 24, 2007

### Gokul43201

Staff Emeritus
At a+b=2, you will not see the equality. There's a lower value of a+b that will realize this.

5. Dec 24, 2007

### Gokul43201

Staff Emeritus
I think not (it's poorly written).

I think they want you to show : 1/(b+c) < 1/(a+c) + 1/(a+b)

6. Dec 24, 2007

### ehrenfest

How is that different? The problem is symmetric in a,b,and c; they are just the lengths of the three sides of a triangle.

What should the problem be then?

Last edited: Dec 24, 2007
7. Dec 24, 2007

### Gokul43201

Staff Emeritus
Doh! Right - I can't remember what I was thinking.

Several possibilities. For one, a+b=\sqrt{2}, everything else unchanged.

8. Dec 24, 2007

### ehrenfest

For 1.8.5, with Gokul's correction, I still don't see how to prove it. I find that the inequality is equivalent to:

$$2 \leq a^2+b^2+ab(y/x +x/y)$$

y/x+x/y is greater than or equal to 2, but I do not see how that helps.

For 1.8.6, I found that the inequality is equivalent to:

$$b^2 < a^2 + b(a+c)+ac+c^2$$

which I am not sure how to prove.