MHB Polynomial inequality

[lfdahl]

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 $P(2) \ge 3^{n}$

 

[Opalg]

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.

 

[lfdahl]

Hint:

Spoiler

Consider the sign of the roots ... - and then take a closer look at $P(2)$

 

[kaliprasad]

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

Spoiler

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 $P(2) >= 3^n$

 

[Opalg]

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 $1$.

 

[lfdahl]

I do not see why Opalg mentioned that coefficients should be integers

Spoiler

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!

 

[Albert1]

I do not see why Opalg mentioned that coefficients should be integers

Spoiler

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 $P(2)>=3^n---(*)$

Spoiler

typo
(*)should be
$P(2) \geq 3^n \sqrt[3]{\prod_{k=1}^{n} (a_k)} = 3^n$ and hence $P(2) \geq 3^n$

 

[kaliprasad]

Spoiler

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 $P(2)>=3^n---(*)$

Spoiler

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