quadratic inequalities

JOANODEA

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

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

JOANODEA

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

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

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

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

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

JOANODEA wrote: »

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

Indeed!

JOANODEA

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

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

Brilliant. Many thanks.

JOANODEA

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

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

JOANODEA

Apologies Jane. :mad:

LeixlipRed

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

equivariant

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

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

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

LeixlipRed

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

Sorry about that, will do in future.

eoins23456

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