I want inequality questons (1 Viewer)

[Richard Lee]

Maths Ext2

I need more inequality question!

 

[ND]

Try this:

a/sqrt(a^2+8bc) + b/sqrt(b^2+8ca) + c/sqrt(c^2+8ab) >= 1

 

[Affinity]

try this:

a, b> 0
a+b <1

prove (a+1)(b+1)< 1/[1-(a+b)]

 

[Affinity]

ND, do you have an elegant solution to your problem by any chance? me can't find it :S

EDIT: Turtle just told me where that one came from.. no wonder you're grinning " "

 

[ND]

http://imo.wolfram.com/problemset/IMO2001_solution2.html

Heheheh.

 

[Richard Lee]

Thanx!

 

[Richard Lee]

To Affinity:
[1+(a+b)+ab][1-(a+b)]=1-(a+b)^2+ab-ab(a+b)=1-a^2-ab-b^2-a^2b-ab^2<1
Therefore:
(a+1)(b+1)<1/[1-(a+b)]


Could u offer more?
Thank!

 

[Fosweb]

These are from arnold sample paper 1 in the back:

1. Show a + b >= 2*sqrt(ab)

2. Hence show:
(a+b)(b+c)(c+a) >= 8abc (this one is ok, but the next one i am stuck on)

a/b + b/c + c/d + d/a >= 4


This is what I have got, is this going down the right track or not?
After showing the first bit and rearranging to get a/b:
a/b >= [(2sqrt(ab))/b] - 1 (and similarly for b/c, c/d, d/a)

a/b + b/c + c/d + d/a >= (2sqrt(ab))/b) + (2sqrt(bc))/c) + (2sqrt(cd))/d) + (2sqrt(da))/a) - 4

Where to from there? Are you allowed to put (a+b) etc back into where you have 2sqrt(ab)'s etc?

 

[Affinity]

Originally posted by Richard Lee
To Affinity:
[1+(a+b)+ab][1-(a+b)]<[1+(a+b][1-(a+b)]
Thank!

excuse my ignorance.. I don't see how that's correct.
____________________________

Fosweb:
see if this helps:
prove (a/b + b/c) + (c/d + d/a) >= 4
you will use the given result 2 times exactly

 

[turtle_2468]

I think he's a fraud. Or at least quite bad at maths. and seeing he claims to be a tutor..

 

[Affinity]

I second that.

 

[McLake]

Hmm, watching thread ...

 

[Richard Lee]

The solution:
(a/b+b/c+c/d+d/a)/4>=4th sqrt[(a/b)(b/c)(c/d)(d/a)]=4
Same as:
(a1+a2+...+an)/n>=nth sqrt(a1*a2*...*an)

 

[Richard Lee]

To turtle:
This is a forum and everybody want to get help from others. I wish that we will be friends. Not like this!
Thanx!

 

[Richard Lee]

Any question? Go to thread " Any hard question"

 

[freaking_out]

Originally posted by ND
Try this:

a/sqrt(a^2+8bc) + b/sqrt(b^2+8ca) + c/sqrt(c^2+8ab) >= 1

i know this is from the olympiad...but is it possible that this can appear in the hsc??

 

[Archman]

Originally posted by Richard Lee
TO ND:
first, prove sqrt(a^2 + 8bc)=<sqrt(a^2+b^2+c^2)
sqrt(b^2+8ac)=<sqrt(a^2+b^2+c^2)
sqrt(c^2+8ab)=<sqrt(a^2+b^2+c^2)
Then prove (a+b+c)>=sqrt(a^2+b^2+c^2)
Therefore:
a/sqrt(a^2+8bc)+b/sqrt(b^2+8ac)+c/sqrt(c^2+8ab)>=1

Could u offer more?
Thank!

i think the stupid html code altered his solution, he meant:
"first, prove sqrt(a^2 + 8bc) =< sqrt(a^2+b^2+c^2)
sqrt(b^2+8ac) =< sqrt(a^2+b^2+c^2)
sqrt(c^2+8ab) =< sqrt(a^2+b^2+c^2)
Then prove (a+b+c) >= sqrt(a^2+b^2+c^2)
Therefore:
a/sqrt(a^2+8bc)+b/sqrt(b^2+8ac)+c/sqrt(c^2+8ab) >= 1"

but the proof is still not right, look at the first line for example:
sqrt(a^2 + 8bc) =< sqrt(a^2+b^2+c^2)
if you throw in a=b=c=1, you get sqrt 9 =< sqrt 3 which isn't true.

 

[turtle_2468]

no there's no chance it'll appear... trust me

 

[freaking_out]

Originally posted by turtle_2468
no there's no chance it'll appear... trust me

well i've been hearing that this years papers gonna b extra hard, i mean drbuchanan was showing us "riemens hypothesis" bcoz he speculates that its gonna be in the exam.

edit: i have no bloody idea what riemens hypothesis is either.:mad1:

 

[turtle_2468]

no chance that'll appear either
You know hard?
Well, this q is around 5 degrees above it.
If you REALLY want the proof, pm me and I'll give it to you...
That's what they say every year btw... that it's going to be extra hard... but the point of the paper is to distinguish the candidates anyway, so they can't really make anything except q8 that much harder
and... who cares? The point is to get the best ranking possible right... so if there's an ultra hard q that u can't do, it's ok as long as others can't do it either!