Solving 1.8.5 Part b): Is It a Counterexample?

[ehrenfest]

Homework Statement

Is what I wrote on the left hand margin a counterexample to 1.8.5 part a) ?

EDIT: I meant part b)

Homework Equations

The Attempt at a Solution

 

[Shooting Star]

Yes. (For b).

 

[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) ?

 

[Gokul43201]

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

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

 

[Gokul43201]

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

I think not (it's poorly written).

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

 

[ehrenfest]

I think not (it's poorly written).

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

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.

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

What should the problem be then?

 

[Gokul43201]

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.

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

What should the problem be then?

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

 

[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.