Prove a_0+a_1+…+a_2016>3^(2017)−1

[lfdahl]

Suppose, that the polynomial $P(x) = x^{2017}+a_{2016}x^{2016}+ a_{2015}x^{2015}+ … + a_1x + a_0$ has $2017$ real roots,

while the polynomial $P(Q(x))$, where $Q(x) = \frac{1}{4}x^2+x-1$, has no real root.

Prove, that $a_0 + a_1 + … + a_{2016} > 3^{2017}-1.$

 

[Opalg]

Suppose, that the polynomial $P(x) = x^{2017}+a_{2016}x^{2016}+ a_{2015}x^{2015}+ … + a_1x + a_0$ has $2017$ real roots,

while the polynomial $P(Q(x))$, where $Q(x) = \frac{1}{4}x^2+x-1$, has no real root.

Prove, that $a_0 + a_1 + … + a_{2016} > 3^{2017}-1.$

[sp]If $\lambda_1,\,\lambda_2,\ldots,\lambda_{2017}$ are the real roots of $P(x)$ then $P(x) = (x - \lambda_1)(x - \lambda_2)\cdots (x - \lambda_{2017}).$

If there is a real number $x$ such that $Q(x) = \lambda_k$ for some $k$, then $x$ would be a real root of $P(Q(x))$. Therefore none of the roots $\lambda_k$ can be in the range of the polynomial $Q(x)$.

The minimum value of $Q(x)$ is $Q(-2) = -2$, so the range of $Q(x)$ is $[-2,\infty)$. Therefore $\lambda_k < -2$ for all $k$. It follows that $$\begin{aligned}P(1) &= (1 - \lambda_1)(1 - \lambda_2)\cdots (1 - \lambda_{2017}) \\ &> (1 - (-2))(1 - (-2))\cdots (1 - (-2)) = 3^{2017}.\end{aligned}$$ But $P(1) = 1+a_{2016} + a_{2015} + \ldots + a_1 + a_0$. Thus $a_0 + a_1 + \ldots + a_{2016} > 3^{2017}-1.$
[/sp]

 

[lfdahl]

[sp]If $\lambda_1,\,\lambda_2,\ldots,\lambda_{2017}$ are the real roots of $P(x)$ then $P(x) = (x - \lambda_1)(x - \lambda_2)\cdots (x - \lambda_{2017}).$

If there is a real number $x$ such that $Q(x) = \lambda_k$ for some $k$, then $x$ would be a real root of $P(Q(x))$. Therefore none of the roots $\lambda_k$ can be in the range of the polynomial $Q(x)$.

The minimum value of $Q(x)$ is $Q(-2) = -2$, so the range of $Q(x)$ is $[-2,\infty)$. Therefore $\lambda_k < -2$ for all $k$. It follows that $$\begin{aligned}P(1) &= (1 - \lambda_1)(1 - \lambda_2)\cdots (1 - \lambda_{2017}) \\ &> (1 - (-2))(1 - (-2))\cdots (1 - (-2)) = 3^{2017}.\end{aligned}$$ But $P(1) = 1+a_{2016} + a_{2015} + \ldots + a_1 + a_0$. Thus $a_0 + a_1 + \ldots + a_{2016} > 3^{2017}-1.$
[/sp]

The ink has hardly dried, before you came up with this excellent solution, Opalg!
Thankyou very much for your participation!(Clapping)