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

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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

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

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since the first way gives us what we want
The far left is 42, not 48 as we want.
yep did you read my recent post
Yes, I do. Thank you so much. ha!! I don't think it is that simple. hehehe ...
why not?
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.
but why don't you think my way works?
I know you heard of the triangle inequality. That is what I used above.
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."
@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.
alright I just did misunderstood above
I know how to use triangle inequality. I just don't think of it on this problem.
Overestimate the problems is my "disease"
i think we all do that

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