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 $y^2={4x}{(1-x)^2}$ 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said, today "routine" means something completely different and I just retreat in shame from the brave new world with my outdated language and ideas...