Home » Ask & Discuss » Mathematics. » Algebra « Back to Discussion

Algebra

Blazing goIITian

Joined:	13 Aug 2008
Post:	1313

9 Nov 2008 16:29:16 IST

0 People liked this

1340

Today's RMO Question --

None

Suppose 'a' and 'b' are real numbers

such that the roots of the cubic equation

ax³ - x² + bx - 1 = 0

are all positive real numbers.

Prove that (i) 0 < 3ab <= 1

(ii) b >= root(3)

pls note that <= and >= stand 4 less than/ equal to and gr8er than/ equal to respectively..

Share this article on:

Comments (13)

Mirka

Blazing goIITian

Joined: 13 Aug 2008

Posts: 1313

9 Nov 2008 16:32:23 IST

Like 0 people liked this

sorry 4 d wide spacing....

this sum i took roots as alpha, beta and gamma

and proved that a and b > o

then tried to use

A.M >= G.M >= H.M

but got stuck with the rearranging of terms...

Can anybody solve this??

santhosh in NUS

Hot goIITian

Joined: 13 Aug 2007

Posts: 132

9 Nov 2008 18:06:51 IST

Like 1 people liked this

Re:Today's RMO Question --

santhosh in NUS

Hot goIITian

Joined: 13 Aug 2007

Posts: 132

9 Nov 2008 18:09:16 IST

Like 3 people liked this

Re:Today's RMO Question --

Ashutosh Sharma

Blazing goIITian

Joined: 29 Sep 2008

Posts: 628

9 Nov 2008 18:16:54 IST

Like 3 people liked this

Using the AM-GM Relations, and

All are positive, so dividing.

As fer the first part, using AM-GM, we wud get and from the above inequality,

Dividing, Hence, ......And as roots are +ve, sum n product nd sum taken 2 at a time r +ve i.e. a,b are +ve. Hence ab>03ab>o

So,

Rahul Duggal

Scorching goIITian

Joined: 10 May 2008

Posts: 289

9 Nov 2008 18:51:09 IST

Like 2 people liked this

Re:Today's RMO Question --

Decoder

Blazing goIITian

Joined: 1 Apr 2007

Posts: 1084

9 Nov 2008 19:05:39 IST

Like 2 people liked this

this ques i have solved by algebra as well as calculus...easiest of the lot..after the first one..wic was a bit easier than this...

look..

by a.m-g.m 1/a more than 3root3...

by am -hm .. b more than 9 / 3root3..

so b more than root3...

actually my shift key isn't working..

Rajat Sen

Blazing goIITian

Joined: 7 Aug 2007

Posts: 533

9 Nov 2008 20:21:56 IST

Like 3 people liked this

let the positive real roots be : $p$ , $q$ , $r$

so we get :

$p+q+r = \frac{1}{a}$

$pq+qr+pr=\frac{b}{a}$

$pqr=\frac{1}{a}$

now $3ab=3\frac{pq+qr+pr}{(p+q+r)^{2}}$

now this is obviously greater than 0.

now we write $pq+qr+rp=\frac{1}{2}((p+q+r)^{2}-(p^{2}+q^{2}+r^{2}))$

this implies $3ab=\frac{3}{2}\left(1-\frac{(p^{2}+q^{2}+r^{2})}{(p+q+r)^{2}}\right)$

now apply power-mean to get $3ab \leq \frac{3}{2}\left(1-\frac{1}{3}\right)$

this gives us the required inequality :

$0<3ab \leq 1$

sending the second part in a while

Rajat Sen

Blazing goIITian

Joined: 7 Aug 2007

Posts: 533

9 Nov 2008 20:44:28 IST

Like 5 people liked this

now the second part is even easier :

now carrying on from above we see that $b=\frac{1}{p}+\frac{1}{q}+\frac{1}{r}$

but as $pqr=p+q+r$

so we get $\frac{1}{pq}+\frac{1}{qr}+\frac{1}{pr}=1$

now from cauchy schawrz we get :

$\left(\frac{1}{p^{2}}+\frac{1}{q^{2}}+\frac{1}{r^{2}}\right)^{2} \geq 1$

so :

$b^{2}-2\left(\frac{1}{pq}+\frac{1}{qr}+\frac{1}{pr}\right) \geq 1$

$\Rightarrow b^{2} \geq 3$

now as b is positive so we can safely right : $b \geq \sqrt{3}$

rate if useful! Idea

Dipanjan

Blazing goIITian

Joined: 30 Jul 2008

Posts: 335

9 Nov 2008 20:59:51 IST

Like 4 people liked this

The first part I have done exactly as rajat has done. My solution to the second part is as follows.

see that $b=\frac{pq+qr+rp}{p+q+r}=\frac{pq+qr+rp}{pqr}$

so, $b^2=\frac{(pq+qr+rp)^2}{pqr(p+q+r)}$

or, $b^2=\frac{(pq+qr+rp)\left(\frac{1}{p}+\frac{1}{q}+\frac{1}{r}\right)}{(p+q+r)}$

$=\frac{2(p+q+r)+\frac{pq}{r}+\frac{qr}{p}+\frac{rp}{q}}{p+q+r}$

$=2+\frac{\frac{pq}{r}+\frac{qr}{p}+\frac{rp}{q}}{p+q+r}$

if we assume $p\ge q\ge r$

then $pq \ge rp \ge qr$

and $\frac{1}{r}\ge \frac{1}{q}\ge \frac{1}{p}$

By Chebychev's ineq.:

$\frac{\frac{pq}{r}+\frac{qr}{p}+\frac{rp}{q}}{3}\ge\frac{(pq+qr+rp)}{3}.\frac{\left(\frac{1}{r}+ \frac{1}{q}+ \frac{1}{p}\right)}{3}$

or, $\frac{\frac{pq}{r}+\frac{qr}{p}+\frac{rp}{q}}{p+q+r}\ge\frac{(pq+qr+rp)}{p+q+r}.\frac{\frac{1}{r}+ \frac{1}{q}+ \frac{1}{p}}{3}=\frac{1}{3}b^2$

or, $b^2\ge2+\frac{1}{3}b^2$

or, $b\ge\sqrt{3}$

Ady JEE-09 AIR 294

Hot goIITian

Joined: 6 Nov 2008

Posts: 145

9 Nov 2008 21:08:07 IST

Like 2 people liked this

differentiate the given equation u get 3ax^2 - 2x + b = 0

as the roots of the cubic are all positive, so will the roots of its derivative be positive as the roots of the derivative lie between the roots of the original cubic

af(k)>0 for k to lie outside the roots

3a(b)>0

as k = 0

hence we have 3ab>0

now D>=0

4 - 4X3ab>=0

giving 3ab<=1

still working on the 2nd part for a shorter solution

rate if correct

Rahul Duggal

Scorching goIITian

Joined: 10 May 2008

Posts: 289

9 Nov 2008 22:04:49 IST

Like 1 people liked this

I am not sure of my solution. Please correct me if I am wrong.

Ady JEE-09 AIR 294

Hot goIITian

Joined: 6 Nov 2008

Posts: 145

9 Nov 2008 22:30:33 IST

Like 1 people liked this

i suppose am>=gm is the only way to get b>=root(3)

Hari Shankar

Forum Expert

Joined: 28 Feb 2007

Posts: 2173

12 Nov 2008 09:18:11 IST

Like 3 people liked this

From the relations given you get $b = \frac{pq+pr+qr}{pqr} = \frac{1}{p}+\frac{1}{q}+\frac{1}{r}$

Also, $\frac{1}{pq}+\frac{1}{qr}+\frac{1}{rp} = 1$

So, we are required to prove that $\left(\frac{1}{p} + \frac{1}{q} +\frac{1}{r} \right)^2 \ge 3\left( \frac{1}{pq}+\frac{1}{qr}+\frac{1}{rp} \right)$

which is analogous to the well known inequality $(a+b+c)^2 \ge 3(ab+bc+ca)$

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team