quadratic inequalities

JOANODEA · 21-10-2010 10:17PM #1

If a squared + b squared = 1 = c squared + d squared, show that ab + cd is
less than or equal to 1 for all a, b, c, d E R.

Appreciate any help.
Thanks.

MathsManiac · 21-10-2010 11:51PM

Start by expanding (a-b)^2. Try to show that ab <= 1/2.

JOANODEA · 22-10-2010 12:08AM

MathsManiac wrote: »

Start by expanding (a-b)^2. Try to show that ab <= 1/2.

Right. (a-b)^2 = a^2 - 2ab +b^2 = 1 - 2ab
and therefore .....

Am I on right track here?

MathsManiac · 22-10-2010 12:14AM

Yep, and now try to use the fact that (a-b)^2 can't be negative, as it's the square of a real number.

equivariant · 22-10-2010 01:30PM

JOANODEA wrote: »

If a squared + b squared = 1 = c squared + d squared, show that ab + cd is
less than or equal to 1 for all a, b, c, d E R.

Appreciate any help.
Thanks.

By the way, are you sure you have transcribed the question correctly? Should it read "... show that ac+bd is less than or equal 1...." (rather than what you typed). Both statements are true, but one is more natural than the other in a geometric sense.

MathsManiac · 22-10-2010 08:48PM

Here's a follow on problem: can you show that the same result is true under the more general condition that a^2 + b^2 + c^2 + d^2 = 2?

JOANODEA · 22-10-2010 10:30PM

MathsManiac wrote: »

Yep, and now try to use the fact that (a-b)^2 can't be negative, as it's the square of a real number.[/QUO

Sorry but can't see how the (a-b)^2 comes into play when its the ab and later the cd we need to sort.
I know I missing something quite blatant.

Hang on a mo.
(a-b)^2 >= 0 which means that 2ab <= 1 which means ab <= 1/2

MathsManiac · 23-10-2010 10:46AM

JOANODEA wrote: »

Hang on a mo.
(a-b)^2 >= 0 which means that 2ab <= 1 which means ab <= 1/2

Indeed!

JOANODEA · 26-10-2010 09:24PM

Hi everyone. Would greatly appreciate help with following problem. Many thanks in advance.

if a, b, c, are positive unequal real nos.
using a^2 + b^2 > 2ab a^2 + c^2 > 2ac b^2 + c^2 > 2bc

show that
3(a^3 + b^3 + c^3) > (a^2 + b^2 + c^2)(a + b + c)

MathsManiac · 27-10-2010 08:29AM

Tricky one, unless I've missed something obvious.

Factorise a^3+b^3 and hence show that it's bigger than ab(a+b).
Do something similar for b&c and for c&a.
Add the three inequalities and you're nearly there.

(It might also help to multiply out the right-hand side of the thing you're trying to prove.)

JOANODEA · 27-10-2010 03:25PM

Brilliant. Many thanks.

JOANODEA · 27-10-2010 03:29PM

A chairde. Would appreciate some help please.

p, q, r, s > or = 0

prove that

(i) (p + q)(q + r)(r + p) > or = 8pqr

and

(ii) (pq + rs)(pr + qs) > or = 4pqrs

LeixlipRed · 27-10-2010 03:52PM

Jane, could you just use the one thread for all these queries please? Thanks.

JOANODEA · 27-10-2010 05:01PM

Apologies Jane. :mad:

LeixlipRed · 27-10-2010 06:23PM

No, it's ok. Just easier to keep it all together.

equivariant · 27-10-2010 10:04PM

All of these inequalities seem to be related (at least conceptually) to the rearrangement inequality

http://en.wikipedia.org/wiki/Rearrangement_inequality

For example

(p+q)(q+r)(r+p) >= 8pqr if and only if
(pq^2+pr^2)+(qr^2+qp^2)+(rp^2+rq^2) >= 6pqr (to see this, multiply out the LHS and cancel some terms)

Now compare pq^2+pr^2 with 2pqr (using rearrangement or otherwise) and then it drops out.

jas2002 · 28-10-2010 02:51AM

For (p + q)(q + r)(r + p) > or = 8pqr

........... I think this is right!

(p+q)(q+r)(r+p)

(pr + pq +q^2 +qr) (r + p)

pr^2 +pqr + rq^2 + qr^2 +qp^2 + pq^2 +pqr +rp^2

2pqr + q^2r + p^2r + p^2q + r^2q + q^2p + r^2p

Balance the eqtn (or should I say statement!) by adding 3x +2pqr and 3x -2pqr, in readiness for squares

2pqr + q^2r -2pqr + p^2r + 2pqr + p^2q -2pqr r^2q +2pqr + q^2p -2pqr + r^2p +2pqr

2pqr + r(q^2 -2pq+ p^2) + 2pqr + q(p^2 -2pr + r^2) +2pqr +

p(q^2 -2qr + r^2) + 2pqr

8pqr + r(q^2 -2pq+ p^2) + q(p^2 -2pr + r^2) + p(q^2 -2qr + r^2)

8pqr + r(q-p)^2 +q(p-r)^2 + p(q-r)^2

Since the squares will be >= 0, then the whole lot will be >= 8pqr (least all are =0)

.........................
The other one is done similarly by adding 2x -2pqrs and 2x +2pqrs to complete the squares. PM if you have a prob with this J.

JOANODEA · 28-10-2010 01:47PM

You are so kind. Worked aa treat. Mile buiochas. Slan go foill.

LeixlipRed · 28-10-2010 03:32PM

jas2002 wrote: »

For (p + q)(q + r)(r + p) > or = 8pqr

........... I think this is right!

(p+q)(q+r)(r+p)

(pr + pq +q^2 +qr) (r + p)

pr^2 +pqr + rq^2 + qr^2 +qp^2 + pq^2 +pqr +rp^2

2pqr + q^2r + p^2r + p^2q + r^2q + q^2p + r^2p

Balance the eqtn (or should I say statement!) by adding 3x +2pqr and 3x -2pqr, in readiness for squares

2pqr + q^2r -2pqr + p^2r + 2pqr + p^2q -2pqr r^2q +2pqr + q^2p -2pqr + r^2p +2pqr

2pqr + r(q^2 -2pq+ p^2) + 2pqr + q(p^2 -2pr + r^2) +2pqr +

p(q^2 -2qr + r^2) + 2pqr

8pqr + r(q^2 -2pq+ p^2) + q(p^2 -2pr + r^2) + p(q^2 -2qr + r^2)

8pqr + r(q-p)^2 +q(p-r)^2 + p(q-r)^2

Since the squares will be >= 0, then the whole lot will be >= 8pqr (least all are =0)

.........................
The other one is done similarly by adding 2x -2pqrs and 2x +2pqrs to complete the squares. PM if you have a prob with this J.

Read the charter, we're not here to do questions for people, we're here to point them in the right direction.

jas2002 · 28-10-2010 05:14PM

Sorry about that, will do in future.

eoins23456 · 02-11-2010 02:17PM

http://www.mathsireland.com/LCHGeneralNotes/Algebra/Proofs/mom_QuesB.html

quadratic inequalities

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