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, $Q(k,x) &gt;0$ for any real number $x$).
  Prove or disprove.
$x\mapsto Q(5, x)$ 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 $Q(L, x_0) &gt; 0$ then $x\mapsto Q(L+\epsilon, x)$ is positive for $\epsilon &gt; 0$ small enough, which contradict the definition of $L$. If $Q(L, x_0) &lt; 0$, then $Q(L-\epsilon, x_0)$ is stille negative for $\epsilon &gt; 0$ 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));
You can try to prove your assumption using induction :

$1.$ Show that :

$Q(5,x)=27\cdot x^6-720 \cdot x^4+2000 \cdot x^3+6000\cdot x^2-24000\cdot x+24000 &gt; 0$  ,for all $x$

$2.$ Suppose that :

$Q(k,x)=27x^6 - 144kx^4 + 80k^2x^3 + 240k^2x^2 - 192k^3x + (64k^4 - 128k^3) &gt;0$ ,for all $x$

$3.$ Try to prove that : $Q(k+1,x) &gt;0 $  , for all $x$ using assumption from step $2$