Showing that complicated mixed polynomial is always positive I want to show that
$\left(132 q^3-175 q^4+73 q^5-\frac{39 q^6}{4}\right)+\left(-144 q^2+12 q^3+70 q^4-19 q^5\right) r+\left(80 q+200 q^2-243 q^3+100 q^4-\frac{31 q^5}{2}\right) r^2+\left(-208 q+116 q^2+24 q^3-13 q^4\right) r^3+\left(80-44 q-44 q^2+34 q^3-\frac{23 q^4}{4}\right) r^4$
is strictly positive whenever $q \in (0,1)$ (numerically, this holds for all $r \in \mathbb{R}$, although I'm only interested in $r \in (0,1)$).
Is that even possible analytically? Any idea towards a proof would be greatly appreciated. Many thanks!

EDIT: Here is some more information.
Let
$f(r) = A + Br + Cr^2 + Dr^3 + Er^4$
be the function as defined above.
Then it holds that
$f(r)$ is a strictly convex function in $r$ for $q \in (0,1)$, $f(0) > 0$, $f'(0) < 0$, and $f'(q) > 0$. Hence, for the relevant $q \in (0,1)$, $f(r)$ attains its minimum for  some $r^{min} \in (0,q)$.
$A$ is positive and strictly increasing in $q$ for the relevant $q \in (0,1)$,
$B$ is negative and strictly decreasing in $q$ for the relevant $q \in (0,1)$,
$C$ is positive and strictly increasing in $q$ for the relevant $q \in (0,1)$,
$D$ is negative and non-monotonic in $q$, and
$E$ is positive and strictly decreasing in $q$ for the relevant $q \in (0,1)$.
 A: Because you want to show that this is always positive, consider what happens when $q$ and $r$ get really big. The polynomials with the largest powers will dominate the result.
You can solve this quite easily by approximating the final value using a large number of inequalities.
A: I would do a mixture of numerical and analytic approaches. I'll consider $r<1$ only.
First, note that an increasing geometric progression has positive finite differences of all orders and those differences form a geometric progression themselves. Thus, it makes sense to decompose the geometric progressions into the elementary polynomials of the form $1$, $x$, $x(x-1)/2$, $x(x-1)(x-2)/6$, etc. (only starting from the right end because our progressions are decreasing):
$$
(1,r,r^2,r^3,r^4)=r^4(1,1,1,1,1)+r^3(1-r)(4,3,2,1,0)+r^2(1-r)^2(6,3,1,0,0)+\dots
$$ 
and similarly for $q$.
The program below (in Asymptote) does the decomposition. I'm too lazy to rewrite the output, but if you run it and look at the resulting $5\times 7$ matrix, you'll see with a naked eye that the diagonals going left bottom to right top define non-negative polynomials, which is the end of the story. 

real[][] C=
{
{0,0,0,132,-175,73,-39/4},
{0,0,-144,12,70,-19,0},
{0,80,200,-243,100,-31/2,0},
{0,-208,116,24,-13,0,0},
{80,-44,-44,34,-23/4,0,0}
};

real[][] X=
{
{1,1,1,1,1},
{4,3,2,1,0},
{6,3,1,0,0},
{4,1,0,0,0},
{1,0,0,0,0},
};

real[][] Y=
{
{1,1,1,1,1,1,1},
{6,5,4,3,2,1,0},
{15,10,6,3,1,0,0},
{20,10,4,1,0,0,0},
{15,5,1,0,0,0,0},
{6,1,0,0,0,0,0},
{1,0,0,0,0,0,0},
};

real[][] Z=X*C*transpose(Y);
write(Z);

pause();

