Define $P(x)$ to be positive if $P(x)>0$ for $x>0$.


I can prove that a quadratic positive polynomial is the ratio of 2 polynomials with non negative coefficients, for example $\displaystyle x^2-x+1/3=\frac{x^6+1/27}{x^4+x^3+2/3 x^2+1/3 x+1/9}$, and similarly for every $x^2-x+c$ where $c>1/4$. The full proof is not hard and involves some recursive polynomials related to Chebyshev polynomials of the second kind.


This leads me to wonder


**QUESTION**: in general is it true that $P$ positive $\Leftrightarrow$ $\exists Q,R$ with non-negative coefficients, such that $\displaystyle P=Q/R$?



Is every positive polynomial the ratio of 2 positive coefficient polynomials?
It is well-known. It is even known that you may take $R=(1+x)^m$ for large enough $m$. See, for example, [John Scholes's solution to Problem 11 of the 38th IMO 1997 shortlist](https://prase.cz/kalva/short/soln/sh9711.html). Note that by the real fundamental theorem of algebra, the general case reduces to the case $\deg P\leqslant 2$.


