Prove that all the roots of $p(x)=F_{n}x^{n}+..+F_{1}x+F_{0}$ can't be real Last night I have created this problem.
Let $p(x)=F_{n}x^{n}+..+F_{1}x+F_{0}$ where $F_{n}$ is $n$th Fibonacci number. Prove that all the roots of $p(x)$ can't be real.  
Edit 1:
$n>1$. 
 A: Let $a_1,a_2,\ldots,a_n$ be the $n$ roots. We then have that
$$\sum a_i^2 = \left(\sum a_i\right)^2- 2\sum_{i \neq j} a_i a_j = \left(\dfrac{F_{n-1}}{F_n} \right)^2 - 2\left(\dfrac{F_{n-2}}{F_n}\right)$$
We also have that
$$\left \vert \prod a_i \right \vert = \dfrac{F_0}{F_n} = \dfrac1{F_n}$$
If all $a_i$'s were to be real, then $a_i^2 \geq 0$ and hence from AM-GM, we have
$$\left(\dfrac{F_{n-1}}{F_n} \right)^2 - 2\left(\dfrac{F_{n-2}}{F_n}\right) \geq \dfrac{n}{F_n^{2/n}}$$
This gives us that
$$F_{n-1}^2 - 2F_n F_{n-2} \geq n F_n^{2-2/n}$$
However, we have that
$$F_{n-1}^2 - 2F_n F_{n-2} = (-1)^{n}$$
Hence, we obtain the contradiction.
A: Thanks @Leg for a nice solution. Here is my own solution
Using Viete's theorem we obtain $\sum \alpha_{i}^{2} = \frac{a_{n-1}^{2}}{a_{n}^{2}}-2 \frac{a_{n-1}}{a_{n}}$ where  $\alpha_{i}$ are the roots of the polynomial.
If all $\alpha_{i}$ is real then  LHS $\geq 0 $ .Since  $a_{i}=F_{i} $ for all $0 \leq i \leq n$ so if all the roots are real then $ \frac{F_{n-1}^{2}}{F_{n}^{2}}-2 \frac{F_{n-1}}{F_{n}} \geq 0$ . But $ \frac{F_{n}}{F_{n-1}}> \frac{F_{n-1}}{F_{n-2}}$ which can be proved with a simple induction argument. Hence we arrive at a contradiction.  

If coefficients are Catalan numbers instead of Fibonacci numbers, then also result remains valid.

