struggling with inequality

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yet2spark

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Sep 28, 2011
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Hi,

I have \(\displaystyle |p^2+2pq-q^2|<|p^2+q^2|\)

I had gone about it like this...

subtract \(\displaystyle p^2\) from both sides \(\displaystyle 2pq-q^2<q^2\)
add \(\displaystyle q^2\) to both sides \(\displaystyle 2pq=2q^2\)
divide both sides by \(\displaystyle 2q\) giving \(\displaystyle p<q\)

but a quick check...

let p=-1 and q=2 (so p < q) I get \(\displaystyle -7 < 5 \) which is correct but \(\displaystyle |-7|\not<|5|\)

what do I need to do to correct this? I know that \(\displaystyle 0<\dfrac{p}{q}<1\) works but can't figure the working to get there
 
JeffM said:
I cannot tell you how to correct "this" because you have not given what the problem says to do.
Click to expand...

Thanks for the tips JeffM.

I want to show that the inequality \(\displaystyle |p^2+2pq-q^2|<|p^2+q^2|\) is satisfied by \(\displaystyle 0<\dfrac{p}{q}<1\) (which it is).
For \(\displaystyle 0<\dfrac{p}{q}<1\ \ \ \ p\ and\ q\) must have the same sign and \(\displaystyle |p|<|q|\)
I just can't figure the working from one to the other. It's the |absolute| value that's muddling me up when trying work it out.
 
Last edited:
(p2 + 2pq - q2)2

= p4 + q4 + 4p2q2 - 2p2q2 + 4p3q - 4q3p

= (p2 + q2)2 - 4pq(q2 - p2)

According to given restrictions → 4pq(q2 - p2) > 0

Then

(p2 + 2pq - q2)2 < (p2 + q2)2 and continue......
 
JeffM said:
As usual, Subhotosh Khan has given an excellent answer. However, because you are having trouble with absolute values, you may have trouble doing the part of the proof that he just stated as "according to the given conditions," which is the heart of the proof. Your proof cannot say "because Subhotosh Khan said so."

lol, whilst grateful to Subhotosh, I was thinking I was back at square one.

The way to do it is prove it first for q < 0 and then for q > 0. (Implicitly q has been defined NOT to equal 0.)

That is frequently a good way to do proofs involving absolute value: You do not know whether the numbers involved are positive, negative, non-positive, or non-negative so your proof must address those cases separately.
Click to expand...

Thanks Jeff, I'll give it a go and let you know if I manage it :)
 
4*p*q*(q2 - p2)

let

A = 4pq

since p and q have same sign → p*q > 0 → A > 0

Now can you show B = (q2 - p2) > 0 from your other given condition?

Then of since A > 0 and B > 0 → A * B > 0
 
No..no...no...

Your answer is different and rigorous (mine is somewhat heuristic) .... it provides a different angle to the proof.... I liked it!!
 
Thanks Jeff and Subhotosh :)

Obviously more work to it than I expected, lol.

It's just that I dislike knowing something is mathematically correct and being unable to prove it.
 
JeffM said:
I have, more times than I like to admit, become convinced that something was mathematically true, but was subsequently unable to prove it. Many times, this has turned out to mean it was not true. (Of course, my personal inability to prove something does not mean it is unprovable, but it has turned out to be a good clue that maybe I should try to find a counter-example.)
Click to expand...

Point taken.

What I'm working on has that many facets and different mathematical connections that, at times, I'm tripping myself up on the simple things. If I finish what I'm attempting I'll let you know. I know that, as in the OP, what I'm asserting is correct, proving it is another matter entirely.

Thanks again
 
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