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Polynomial Inequality Problem

CaptainBlack

Well-known member
Jan 26, 2012
890
I may have posted this back in the Old Country, but:

let the polynomial: \[P(x)=x^n+a_1X^{n-1}+ ... + a_{n-1}x+1 \] have non-negative coeficients and \(n\) real roots.

Prove that \(P(2)\ge 3^n \)

CB
 

Sherlock

Member
Jan 28, 2012
59
A hint would be appreciated.
 

Alexmahone

Active member
Jan 26, 2012
268
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CaptainBlack

Well-known member
Jan 26, 2012
890
A hint would be appreciated.
Will do when I am at home with my notes, but to get you started the non-negativity of the coefficients tell you that there are no positive roots (there is a supprise, CaptainBlack uses Descartes rule of signs). And that there are \(n\) real roots means that:

\[ P(x)=\prod_{i=1}^n(x+a_i),\ \ \ a_i>0 \]

and:

\[ \prod_{i=1}^n a_i=1 \]

CB
 
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CaptainBlack

Well-known member
Jan 26, 2012
890
And the only other idea involved is the AM-GM inequality

CB
 
Last edited:

Sherlock

Member
Jan 28, 2012
59
I've sort of a set-up so far. I'm trying to work backwards but I think I'm missing a step.

$\displaystyle 3^n = \bigg[2+\prod_{1 \le k \le n}a_{k}\bigg]^n \le \bigg[2+\bigg(\prod_{1 \le k \le n}a_{k}\bigg)^{\frac{1}{n}}\bigg]^n \le \bigg[2+\frac{1}{n}\sum_{1 \le k \le n}a_{k}\bigg]^n = \bigg[\frac{1}{n}\sum_{1 \le k \le n}(2+a_{k})\bigg]^{n} \ge \prod_{k=1}^{n}(2+a_{k}) = P(2).$

Is there a missing inequality here that fixes this? I'm really no good with inequalities! (Giggle)

---------- Post added at 03:17 AM ---------- Previous post was at 03:13 AM ----------

I think $3^n=(2+1)^n=\sum_{k=0}^n\dbinom{n}{k}2^k$ will be used somewhere.
I've tried that (by creating a double sum, although I've no idea why) but to no avail unfortunately.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
I've sort of a set-up so far. I'm trying to work backwards but I think I'm missing a step.

$\displaystyle 3^n = \bigg[2+\prod_{1 \le k \le n}a_{k}\bigg]^n \le \bigg[2+\bigg(\prod_{1 \le k \le n}a_{k}\bigg)^{\frac{1}{n}}\bigg]^n \le \bigg[2+\frac{1}{n}\sum_{1 \le k \le n}a_{k}\bigg]^n = \bigg[\frac{1}{n}\sum_{1 \le k \le n}(2+a_{k})\bigg]^{n} \ge \prod_{k=1}^{n}(2+a_{k}) = P(2).$

Is there a missing inequality here that fixes this? I'm really no good with inequalities! (Giggle)

---------- Post added at 03:17 AM ---------- Previous post was at 03:13 AM ----------

I've tried that (by creating a double sum, although I've no idea why) but to no avail unfortunately.
Use the AM-GM inequality on each factor of \(P(2)\):

$$ 2+a_i=1+1+a_i \ge 3(1\times 1\times a_i)^{1/3} $$

CB
 
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CaptainBlack

Well-known member
Jan 26, 2012
890
Use the AM-GM inequality on each factor of \(P(2)\):

$$ 2+a_i=1+1+a_i \ge 3(1\times 1\times a_i)^{1/3} $$

CB

Which of course tells us that the more general result holds:

For any positive integer \(k\) we have \( P(k)\ge (k+1)^n \)

CB
 

Albert

Well-known member
Jan 25, 2013
1,225
Let P(x)=(x+b1)(x+b2)(x+b3)------------(x+bn)

if bi=1 (i=1,2,---------------n) then

$P(x)=(x+1)^n$

$ \therefore P(2)=(2+1)^n=3^n$

for some bi and bj (here bi and bj are not equal to 1 ,then bi and bj are

reciprocal to each other (again i,j=1,2,----n )

let bi=k (k>0) ,then bj=$\dfrac {1}{k}$

$ \therefore P(x)=(x+1)^{n-2}(x+k)(x+\dfrac {1}{k})$

$( \because 2k+\dfrac {2}{k}>4) $

$ \therefore P(2)=(2+1)^{n-2}(2+k)(2+\dfrac {1}{k})> 3^n$