Let $P$ be a polynomial with real coefficients such that $P(x) > 0$ if $x > 0$. Prove that there exist polynomials $Q$ and $R$ with nonnegative coefficients such that $P(x) = \frac{Q(x)}{R(x)}$ if $x > 0.$
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Tags: algebra, polynomial, coefficients, IMO Shortlist
01.08.2012 18:42
Call a polynomial "good" if it has all real, non-negative coefficients. We can WLOG assume that $P(x)$ is monic, so it can be factored into the product of linear and quadratic monic polynomials with real coefficients. The linear factors have real roots $<0$, so they are good. All that remains then is to prove the assertion for any quadratic monic $f(x)$ with no positive real roots. The case where both roots are negative is trivial, so we can assume that the two roots of $f(x)$ are non-real. If $f(x)=x^2 + ax +b$, then $b>0$, and by replacing $x$ with $x\sqrt{b}$, we can assume that $b=1$, which means that $a$, which is negative, is greater than $-2$. So it finally remains to prove the assertion for $f(x)$ of the form $x^2 -2xc os\theta +1$, for some $0 < \theta < \pi /2$. Mulitply $f(x)$ by $x^2 +2xc os\theta +1$, which is good, to get $x^4 -2x^2c os2\theta +1$. If $2\theta \ge \pi$, we're done, otherwise multiply by $x^4 +2x^2c os2\theta +1$ to get $x^8 -2x^4c os4\theta +1$, and keep repeating until $2^k\theta \ge \pi$, for some $k$, and we're done.