## Loser66 one year ago Show that for all z on the circle |z|=3 $$\dfrac{1}{120}\leq |\dfrac{1}{z^4-4z^2+3}|\leq \dfrac{1}{48}$$ Please, help

1. Loser66

@ganeshie8

2. myininaya

I was thinking about using triangle inequality $|z|^4-4|z|^2-|3|\le |z^4-4z^2+3| \le |z|^4+4|z|^2+|3| \\ 3^4-4(3)^2-3 \le |z^4-4z^2+3| \le 3^4+4(3)+3$ then flipping things changes the direction of the inequality

3. myininaya

I guess we could have done this instead on the that one side instead: $|a+b+c| >|a|-|b+c|>|a|- \{|b|-|c|\}=|a|-|b|+|c| \\ \text{ instead of } \\ |a+b+c|>|a+b|-|c|>|a|-|b|-|c|$

4. myininaya

since the first way gives us what we want

5. Loser66

The far left is 42, not 48 as we want.

6. myininaya

yep did you read my recent post

7. Loser66

Yes, I do. Thank you so much. ha!! I don't think it is that simple. hehehe ...

8. myininaya

why not?

9. Loser66

We have a similar problem in class and my prof gave us a very complicated explanation for that. I would like to know if there is any shortcut or not. That's why I post it here.

10. myininaya

but why don't you think my way works?

11. myininaya

I know you heard of the triangle inequality. That is what I used above.

12. myininaya

I'm confused when you said "I don't think it is that simple." did you mean to say "I didn't think it would be that simple."

13. Loser66

@myininaya I didn't mean your method is wrong. I apology if I made you think that I don't believe your method. It is perfectly right. I got it.

14. myininaya

alright I just did misunderstood above

15. Loser66

I know how to use triangle inequality. I just don't think of it on this problem.

16. Loser66

Overestimate the problems is my "disease"

17. myininaya

i think we all do that