Ok, I have not found an elegant way to approach this but with some intuitive guesses I have found a counterexample showing the claim to be false. Now clearly, from being relatively prime and nonconstant, , so the first hope for a counter example would be in which case the inequality fails. For this to work, each polynomial must be either a perfect square or perfect cube of a linear term. It is not difficult to show that we cannot sum two perfect cubes to another perfect cube by considering the resulting system of equations of the coefficients.
My solution was constructed by assuming and , and considering the systems of equations from the coefficients. One set of solutions is , giving the counterexample:
So these polynomials, being nonconstant and sharing no roots, satisfy the requirements. However, has three distinct roots, thus which is absurd.