If $|a|\ge 2$,$|b|\ge 2$, $P(x)=x^4-(a+b)x^3+(ab+2) x^2-(a+b)x+1$, prove that all roots of $P(x)$ are real 
Given: $a,b\in \mathbb R$, $|a|\ge 2$,$|b|\ge 2$,  $$P(x)=x^4-(a+b)x^3+(ab+2) x^2-(a+b)x+1,$$
Prove or disprove: all roots of $P(x)$ are real.

From a math contest. The polynomial is reciprocal, but I can't see how to use this on the proof, if that is the case, and I tried to use Descartes' rule of signs, but without success. I'm a little lost on how to address the problem.
Hints or solutions are appreciated. Sorry if this is a duplicate.

Given the helpful hints and suggestions from David Quinn and Alex Francisco, I will try a full solution to my question, for sake of completeness. 
First, by dividing $P(x)=x^4-(a+b)x^3+(ab+2)x^2-(a+b)x+1=0$ by $x^2$ we get
$$x^2-(a+b)x+(ab+2)-(a+b)\frac{1}{x}+\frac{1}{x^2}=0$$
$$\Leftrightarrow x^2+\frac{1}{x^2}-(a+b)(x+\frac{1}{x})+ab+2=0 $$
but, as $\displaystyle (x+\frac{1}{x})^2=x^2+\frac{1}{x^2}+2,$ the previous equation is equivalent to
$$(x+\frac{1}{x})^2-2-(a+b)(x+\frac{1}{x})+ab+2=0$$
$$\Leftrightarrow u^2-(a+b)u+ab=0~~\text{with the substitution}~~u=x+\frac{1}{x},$$
with obvious solution $u=a$ and $u=b$, real values by the assumption.
Now, as $u=x+\frac{1}{x}$, we can solve for $x$ noticing that
$$u=\frac{x^2+1}{x}\Leftrightarrow x^2-ux+1=0$$
Therefore, as $u=a$ and $u=b$ the original polynomial can be expressed by
$$P(x)=(x^2-ax+1)(x^2-bx+1)=0,$$
and the condition $|a|\ge 2$ and $|b|\ge 2$ ensures that the roots of both equations (that are the roots of $P(x)$) will be real numbers, as the discriminant for each second order degree polynomial will be non-negative with this condition.
