Show that this polynomial is positive Consider the following polynomial in two variables :
$$
Q(k,x)=27x^6 - 144kx^4 + 80k^2x^3 + 240k^2x^2 - 192k^3x + (64k^4 - 128k^3)
$$
Then for any integer $k \geq 5$, the polynomial $Q(k,.)$ (in one variable $x$)
seems to be always positive (i.e, $Q(k,x) >0$ for any real number $x$).
  Prove or disprove.
 A: *

*$x\mapsto Q(5, x)$ is positive. There is numerous ways to prove that, more or less algorithmically.

*Assume that there exists a $k$ such that $x\mapsto Q(k,x)$ is not positive, and consider $L$ be the infimum of all real numbers $k \geqslant 5$ such that $x\mapsto Q(k, x)$ is not positive.
I claim that $Q(L,x)$ has a double real root, this is equivalent to say that $L$ is a root of the discriminant of $Q$ w.r.t. the variable $x$.

*The discriminant of $Q$ with respect to $x$ is (computed with Maple) :
$$ \operatorname{disc}_x Q = 92162779488452608 k^{16} (k-4)^4 $$

*The real roots of this discriminant are 0 and 4, thus $L$ cannot be finite, and hence the claim.
Proofs


*

*Point 1 can be proved using Sturm sequences, and you can compute these sequences using Maple :

sturm(subs(k=5, Q),x,-infinity, infinity);
      0

This means that $Q(5, x)$ has no real roots. Since the leading coefficient is positive, the polynomial $Q(5, x)$ is positive whenever $x\in \mathbb R$.


*

*To prove point 2 consider $x_0$ a real numbers such that $Q(L, x)$ is minimal. We have $Q(L, x_0) = 0$, because if $Q(L, x_0) > 0$ then $x\mapsto Q(L+\epsilon, x)$ is positive for $\epsilon > 0$ small enough, which contradict the definition of $L$. If $Q(L, x_0) < 0$, then $Q(L-\epsilon, x_0)$ is stille negative for $\epsilon > 0$ small enough, which is again a contradiction. So $x_0$ is a root of $Q(L,x)$ and of $\partial_x Q(L,x)$, thus it is a double root. The discriminant thing is usual.

*The point 3 is Maple :
factor(discrim(Q, x));
A: You can also use quantifier elimination algorithms. This gives you precisely for which $k$ the polynomial $Q(k,x)$ is positive. For example Mathematica implements such an algorithm :
In[1] := Resolve[ForAll[{x}, Q > 0]]
Out[1] := k < 0 || k > 4

I insist that you can trust this result, it is an exact algorithm not a numerical heuristic.
A: You can try to prove your assumption using induction :
$1.$ Show that :
$Q(5,x)=27\cdot x^6-720 \cdot x^4+2000 \cdot x^3+6000\cdot x^2-24000\cdot x+24000 > 0$  ,for all $x$
$2.$ Suppose that :
$Q(k,x)=27x^6 - 144kx^4 + 80k^2x^3 + 240k^2x^2 - 192k^3x + (64k^4 - 128k^3) >0$ ,for all $x$
$3.$ Try to prove that : $Q(k+1,x) >0 $  , for all $x$ using assumption from step $2$
