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 $y^2={4x}{(1-x)^2}$ 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said, today "routine" means something completely different and I just retreat in shame from the brave new world with my outdated language and ideas... 1 In good old times the word "routine" meant "easily done by a trained person".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).