Polynomial inequality

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Discussion Overview

The discussion revolves around the polynomial inequality involving a polynomial $P(x)$ with non-negative coefficients and its real roots. Participants are tasked with proving that $P(2) \ge 3^{n}$ under certain conditions regarding the coefficients.

Discussion Character

  • Debate/contested

Main Points Raised

  • Some participants assert that the polynomial $P(x) = 1 + a_1x + a_2x^2 + ... + a_{n-1}x^{n-1} + x^n$ must have non-negative integer coefficients to ensure it has $n$ real roots and to validate the inequality $P(2) \ge 3^{n}$.
  • Others question the necessity of integer coefficients, suggesting that non-negative coefficients alone may suffice for the polynomial to have $n$ real roots.
  • A participant acknowledges a previous oversight regarding the coefficients, indicating a potential misunderstanding in the initial conditions of the problem.
  • Another participant expresses gratitude for a solution provided by a peer, indicating some level of resolution or acceptance of that contribution.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on whether the coefficients must be integers or if non-negative coefficients are adequate. The discussion remains unresolved regarding the implications of coefficient types on the polynomial's properties.

Contextual Notes

The discussion highlights the importance of the type of coefficients in relation to the polynomial's roots and the validity of the inequality, but does not resolve the assumptions or definitions involved.

lfdahl
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The polynomial: $P(x) = 1 + a_1x +a_2x^2+...+a_{n-1}x^{n-1}+x^n$

with non-negative integer coefficients has $n$ real roots. Prove, that $P(2) \ge 3^{n}$
 
Last edited:
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lfdahl said:
The polynomial: $P(x) = 1 + a_1x +a_2x^2+...+a_{n-1}x^{n-1}+x^n$

with non-negative coefficients has $n$ real roots. Prove, that $P(2) \ge 3^{n}$
I guess that should be "The polynomial: $P(x) = 1 + a_1x +a_2x^2+...+a_{n-1}x^{n-1}+x^n$ with non-negative integer coefficients has $n$ real roots." Otherwise the result is not true.
 
Hint:

Consider the sign of the roots ... - and then take a closer look at $P(2)$
 
lfdahl said:
The polynomial: $P(x) = 1 + a_1x +a_2x^2+...+a_{n-1}x^{n-1}+x^n$

with non-negative integer coefficients has $n$ real roots. Prove, that $P(2) \ge 3^{n}$

I do not see why Opalg mentioned that coefficients should be integers
Because all coefficients are positive so all n roots are -ve and hence
$P(x) = \prod_{k=1}^{n} (x+ a_k)$ where all $a_k$ are positive
further $\prod_{k=1}^{n} (a_k) = 1$
so $P(2) = \prod_{k=1}^{n}(2+a_k)\cdots(1)$
now taking AM GM between 1,1 $a_k$ we get $(2+a_k) >= 3\sqrt[3]{(a_k)}\cdots(2)$
so from (1) and (2)
$P(2) >= 3^n \sqrt[3]{\prod_{k=1}^{n} (a_k)} = 3^n$ and hence $P(2) >= 3^n$
 
Last edited:
kaliprasad said:
I do not see why Opalg mentioned that coefficients should be integers
You are quite right. I somehow overlooked the fact that the constant term and the coefficient of $x^n$ are both $1$.
 
kaliprasad said:
I do not see why Opalg mentioned that coefficients should be integers
Because all coefficients are positive so all n roots are -ve and hence
$P(x) = \prod_{k=1}^{n} (x+ a_k)$ where all $a_k$ are positive
further $\prod_{k=1}^{n} (a_k) = 1$
so $P(2) = \prod_{k=1}^{n}(2+a_k)\cdots(1)$
now taking AM GM between 1,1 $a_k$ we get $(2+a_k) >= 3\sqrt[3]{(a_k)}\cdots(2)$
so from (1) and (2)
$P(2) >= 3^n \sqrt[]{\prod_{k=1}^{n} (a_k)} = 3^n$ and hence $P(2) >= 3^n$

Thankyou, kaliprasad, for your solution. Well done!
 
kaliprasad said:
I do not see why Opalg mentioned that coefficients should be integers
Because all coefficients are positive so all n roots are -ve and hence
$P(x) = \prod_{k=1}^{n} (x+ a_k)$ where all $a_k$ are positive
further $\prod_{k=1}^{n} (a_k) = 1$
so $P(2) = \prod_{k=1}^{n}(2+a_k)\cdots(1)$
now taking AM GM between 1,1 $a_k$ we get $(2+a_k) >= 3\sqrt[3]{(a_k)}\cdots(2)$
so from (1) and (2)
$P(2)>=3^n\sqrt[]{\prod_{k=1}^n (a_k)}=3^n$ and hence $P(2)>=3^n---(*)$
typo
(*)should be
$P(2) \geq 3^n \sqrt[3]{\prod_{k=1}^{n} (a_k)} = 3^n$ and hence $P(2) \geq 3^n$
 
Albert said:
Because all coefficients are positive so all n roots are -ve and hence
$P(x) = \prod_{k=1}^{n} (x+ a_k)$ where all $a_k$ are positive
further $\prod_{k=1}^{n} (a_k) = 1$
so $P(2) = \prod_{k=1}^{n}(2+a_k)\cdots(1)$
now taking AM GM between 1,1 $a_k$ we get $(2+a_k) >= 3\sqrt[3]{(a_k)}\cdots(2)$
so from (1) and (2)
$P(2)>=3^n\sqrt[]{\prod_{k=1}^n (a_k)}=3^n$ and hence $P(2)>=3^n---(*)$
typo
(*)should be
$P(2) \geq 3^n \sqrt[3]{\prod_{k=1}^{n} (a_k)} = 3^n$ and hence $P(2) \geq 3^n$

I have done the needful
 

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