TITLE: Proving univariate polynomials (defined by sums, binomial coeffs, etc.) are nonnegative: is it 'routine'?
QUESTION [18 upvotes]: My colleagues and I are working on a project related to an old paper of C. Borell and we have boiled it down to the following problem: 
Show, for all integers $1 \leq i \leq k$, that the univariate real polynomial $P(x) = \frac12 \binom{2k}{2i} (1+x)^{2k-2i} + \frac12 Q(x)$ is everywhere nonnegative, where $Q(x) = \sum_{j=i}^k \binom{2k}{2j}(1-x)^{2k-2j}\binom{j}{i}(-4x)^{j-i}$.  
(For what it's worth, Maple recognizes $Q(x)$ as $\binom{2k}{2i} (1-x)^{2k-2i} \text{hypergeom}([-(k-i), -(k-i)+\frac12], [i+\frac12], -4x/(1-x)^2)$.
I'm not asking MO to prove this (although I suppose if someone saw how to do so immediately I wouldn't turn down the answer).  Instead, I'm asking "Is this 'routine'?" in the sense of the word used in the Petkovsek-Wilf-Zeilberger A=B book?  In other words, would Doron Zeilberger say, "Oh yes, just type the following into Maple and it will produce a proof of the claim"?  In other other words, does this question fall into a class of problems known to be decidable?
Of course, for any fixed $i$ and $k$ the problem is 'routine'.  E.g., for $k = 4$, $i = 2$, we have $P(x) = 70x^4-168x^3+804x^2-168x+70$ and furthermore I can coax my computer into proving that's nonnegative.  (Say, by obtaining a sum-of-squares representation like $P(x) = 42(x-1)^4 + 28x^4 + 552x^2+28$.)
I don't know the  "A=B technology" very well: my question is whether it, or any other techniques, can be used to automatically prove these inequalities.  I would also be happy to accept an answer explaining why proving these inequalities is not obviously 'routine' and will require some ingenuity.  
UPDATE:  Doron Zeilberger wrote me an email describing why at least 'half' of this problem is routine: namely, that for $i$ symbolic and $k-i$ numeric the nonnegativity can be proven by computer.  He preferred not to post himself but said I could post his ideas here; I will do so once I get a chance to think about them.

REPLY [16 votes]: In good old times the word "routine" meant "easily done by a trained person". In this sense, the problem is, indeed, routine. Change the notation to $s=j-i$, $p=k-i$. Observe that negative $x$ are not a problem. For positive $x$, denote $y^2=\frac{4x}{(1-x)^2}$ and rewrite the inequality as 
$$
{2(p+i)\choose 2i}(1+y^2)^p+\sum_{s=0}^p (-1)^s{2(p+i) \choose 2(s+i)}{s+i\choose i}y^{2s}\ge 0.
$$
Now start with $i=0$. Then the ugly junk goes away and we get $(1+y^2)^p+\Re[(1+iy)^{2p}]\ge 0$, which is a no-brainer ($|z|^{2p}+\Re[z^{2p}]\ge 0$ for all complex $z$). Now just define $F_0(y)=\Re[(1+iy)^{2p}$, $F_m(y)=y\int_0^yF_{m-1}(t)dt$ and $G_m(y)=\frac{1}{(2m-1)!!}y^{2m}(1+y^2)^p$. The general inequality is equivalent to $G_m+F_m\ge 0$ (my $m$ is your $i$ but, since I used $i$ for the imaginary unit, to use it for the index now would be quite unfortunate). However $y\int_0^y G_{m-1}(t)dt\le y\int_0^y\frac{1}{(2m-3)!!}t^{2m-2}(1+y^2)^p dt=G_m(y)$ ($(-1)!!=1$, of course). 
But, as you said, today "routine" means something completely different and I just retreat in shame from the brave new world with my outdated language and ideas...