a polynomial question Suppose that we have the polynomial
$$
G(\omega)=8\omega^8+32r^2\omega^6+\gamma_5\omega^4+\gamma_3\omega^2+\gamma_1,
$$
where $\gamma_5,\gamma_3,\gamma_1, r$ are real parameters. If possible, I would like to find conditions on the coefficients of $G$ under which $\omega G(\omega)>0$ for all $\omega > 0$. Please help!
 A: This won't be a very satisfying answer, because your quartic is almost completely general: since positivity issues basically don't depend on the leading coefficient except up to sign, the only condition you've effectively required is that the second coefficient be positive. In particular, there are no simple necessary and sufficient conditions, but here are some comments, including some sufficient conditions.
Write $g=8x^4+a^2x^3+bx^2+cx+d$ the associated quartic. We know very little about this: but since $8>0$, we must have $d\geq 0$ to make sure $g>0$ when $x>0$. If $d=0$ then finding the roots of $g$ is again reduced to finding the roots of the cubic $g'/x$, which you can similarly shortcut in certain cases by looking at $c$, or else use the cubic formula on. So now let's assume $d>0$.
Now if $g$ has any roots with $x>0$, its derivative $g'=32 x^3+3a^2x^2+2bx+c$ must have some such roots as well: one less than the number of simple roots and triple roots (which must be even) and one less than twice the number of double roots if there are any. You can apply the cubic formula to $g'$, which is not as freakishly unwieldy as the quartic formula, and then if $g'$ has no positive real roots then you're golden. (In practice the software you will plug a cubic into is not using the cubic formula unless you want exact answers, which won't be important here unless you have some solutions incredibly close to zero, in particular, some very extreme coefficients.)
For a more special case: if $g'(0)>0$, then $g'$ has a negative real root, and then if the discriminant $3456a^2bc-108a^6c-256b^3 -1728c^2$ is negative-which it generally will be if $a^4$ is more than about $30$, and since you have $a^2=32r^2$, if $r$ is of integer scale this will always happen-then the other two roots are complex and you win.
If $g'$ does have some positive real roots, then you have to actually look at $g$. Some shortcuts here: since $a^2/8$ the sum of roots is positive, if your roots are all real then you lose. You can check this in terms of the discriminant, which is complicated enough for a quartic that I won't type it here: see Wikipedia. If there are some complex roots, then you really just have to compute all the roots, and the formula will not give you anything informative: of course you can just write down the enormously complicated expressions for the roots and say "if this is not positive..."
