Polynomial phobia abandoned!

Here's a proof by generalization.

Let $p(x) = x^{17} +a_1x^{16}+ ....+ a_{15}x^2+a_{16}x-1$ such that $a_i \in \mathbb{R}^+$

$\bullet$ As all coefficients are positive, $p(x)\geq 1$ whenever $x \geq 0$

$\implies$ All the roots are Negative real numbers . Let the roots be $-\alpha_1 , -\alpha_2 , ... , -\alpha_{17}$ .

$\bullet$ Thus let $p(x) = (x+\alpha_1 )(x+\alpha_2)...(x+\alpha_{17})$ where $\alpha_i \in \mathbb{R}^+$

$\implies$ For a real number $t$ , $p(t) = (t+\alpha_1)(t+\alpha_2)...(t+\alpha_{17})$

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Now we apply the AM-GM inequality in a cute way whenever $t$ is integer ... $\dfrac{\overbrace{1+1+..+1}^\text{'t' ones}+\alpha_1}{t+1} \geq \sqrt[t+1]{\underbrace{1\cdot1\cdot.... \cdot1}_\text{'t' ones} \cdot \alpha_1}$ $\dfrac{\overbrace{1+1+..+1}^\text{'t' ones}+\alpha_2}{t+1} \geq \sqrt[t+1]{\underbrace{1\cdot1\cdot.... \cdot1}_\text{'t' ones} \cdot \alpha_2}$ $...$ $\dfrac{\overbrace{1+1+..+1}^\text{'t' ones}+\alpha_{17}}{t+1} \geq \sqrt[t+1]{\underbrace{1\cdot1\cdot.... \cdot1}_\text{'t' ones} \cdot \alpha_{17}}$

Taking the product of all, $\dfrac{(t+\alpha_1)(t+\alpha_2)...(t+\alpha_{17})}{(t+1)^{17}} \geq \sqrt[t+1]{\alpha_1\cdot\alpha_2\cdot...\cdot \alpha_{17}}$

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But the product of all the roots is $1$ as given in question.

$\therefore \dfrac{(t+\alpha_1)(t+\alpha_2)...(t+\alpha_{17})}{(t+1)^{17}} \geq 1$

$\therefore (t+\alpha_1)(t+\alpha_2)...(t+\alpha_{17}) \geq (t+1)^{17}$

$\therefore p(t) \geq (t+1)^{17}$

That gives us $p(2) \geq 3^{17}$ , minimum value of $p(2) = 3^{17} = \boxed{129140163}$

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Note - For a polynomial of degree $n$ with specified conditions, $p(t) \geq (t+1)^n$

We can also do this problem by repeatedly using Vieta's formula and then using AM-GM on each coefficient.

For example:

$q(x) =x^n+a_{n-1}x^{n-1} +...+a_2x^2+a_1x+1$

$q(x) = (x+\alpha_1 )(x+\alpha_2)...(x+\alpha_{17})$ where $\alpha_i \in \mathbb{R}^+$

Since all the roots are of same sign(-ve), we can do this for all $r$ (please reason this out yourself).

$\frac { \sum {\text{sum of (products of }\alpha \text{ taken 'r' at a time)} } }{ \left( \begin{matrix} n \\ r \end{matrix} \right) } \ge \sqrt [ \left( \begin{matrix} n \\ r \end{matrix} \right)]{\text{ product of (products of } \alpha \text{ taken 'r' at a time)} }$

$\Rightarrow \frac { { a }_{ r } }{ \left( \begin{matrix} n \\ r \end{matrix} \right) }\ge 1 \\ \Rightarrow { a }_{ r }\ge \left( \begin{matrix} n \\ r \end{matrix} \right)$

The funny part is that the minimum occurs for each and every $a_r$ for the same condition, that is, in this case: all roots= $-1$

Hence, the required polynomial is $(x+1)^{17}$

Which yields $\boxed{3^{17}}$ at $x=2$ .

This is my solution Please tell me where i went wrong