The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

This discussion is now closed.

Check out other Related discussions

Application of Cauchy-Schwarz

Hi,

I've already shown that (x⃗⋅y⃗)2≤(x⃗⋅x⃗)(y⃗⋅y⃗)(\vec{x} \cdot \vec{y})^2 \leq(\vec{x} \cdot \vec{x})(\vec{y} \cdot \vec{y}) with equality holding if and only if x⃗=λy⃗\vec{x}=\lambda \vec{y}.

I'm then told by considering suitable vectors in R4\mathbb{R}^4, show that only one choice of reals x,y & z satisfies 3(x2+y2+z2−4)−2(yz+xy+zx)−4(x+y+z)=03(x^2+y^2+z^2-4) - 2(yz+xy+zx) - 4(x+y+z) = 0 .

I've tried it for a while but I've no clue what vectors i should be considering here...absolutely baffled. Any help would be massively appreciated. Thanks
Reply 1
Avatar for around
around
13
Have you done the first part of the question? It gives you a massive hint as to the second.

Even if you haven't, you can still use the result of the first part to derive an inequality for the second, and then find a condition for the inequality being an equality (basically substitute in)
Reply 2
Avatar for lilman91
lilman91
OP
4
around
Have you done the first part of the question? It gives you a massive hint as to the second.

Even if you haven't, you can still use the result of the first part to derive an inequality for the second, and then find a condition for the inequality being an equality (basically substitute in)


Yeah I've done the first part. that was easy, I just considered x⃗=(x,y,z)\vec{x}=(x,y,z) and y⃗=(z,x,y)\vec{y} = (z,x,y) and it pops out from the inequality..but I cant seem to find the vectors that make this one work. I imagine its two sets of vectors and u add the inequalities together
Reply 3
Avatar for DFranklin
DFranklin
18
It would be helpful if you actually gave us the first part of the question. If you want to have discussions that only make sense to other people doing your course, there are always the university specific forums.

Sorry: private bugbear.
Reply 4
Avatar for around
around
13
sorry, here's a link: http://www.maths.cam.ac.uk/undergrad/pastpapers/2007/Part_IA/PaperIA_1.pdf . iirc it's about question 7.

Ok, well, try substituting the result of the first part of the question into the second, and then play around with it a little bit. When do you have equality for the first part? What does this imply about (x,y,z)?
Reply 5
Avatar for nuodai
nuodai
17
I'm actually quite interested in how you do this (so please let me know!). I did this question a couple of months ago and did this part of the question by scratching my head for a couple of hours and trying a few things out (which isn't good exam technique).
Reply 6
Avatar for lilman91
lilman91
OP
4
around
sorry, here's a link: http://www.maths.cam.ac.uk/undergrad/pastpapers/2007/Part_IA/PaperIA_1.pdf . iirc it's about question 7.

Ok, well, try substituting the result of the first part of the question into the second, and then play around with it a little bit. When do you have equality for the first part? What does this imply about (x,y,z)?


nuodai
I'm actually quite interested in how you do this (so please let me know!). I did this question a couple of months ago and did this part of the question by scratching my head for a couple of hours and trying a few things out (which isn't good exam technique).


Yeah it was late at night and I feel i was just being stupid. I came to it this mornign and got it instantly.

The expression is equal to:

[2(x2+y2+z2)−2(xy+yz+xz)]+[(x−2)2+(y−2)2+(z−2)2][2(x^2+y^2+z^2) -2(xy+yz+xz)] + [(x-2)^2 +(y-2)^2 + (z-2)^2] .

From the first part of the question, the first of the bracketed terms is ≥0\geq 0, and so is the last part as it is jus squares. So for the equality to hold, we require both terms to be 0

For the first part to be 0, we need x=λz,y=λx,z=λy⇒x=y=z x= \lambda z, y=\lambda x, z=\lambda y \Rightarrow x=y=z and for the second part, we plainly need x=y=z=2x=y=z=2. So that is the only solution!

Tadah
Reply 7
Avatar for around
around
13
I think the method that was hinted at in the question was to use Cauchy-Schwarz on (x,y,z,2) and (1,1,1,1).

But hey, any method works.

Related discussions

Latest

Trending

Trending

Articles for you