How do we prove $x^6+x^5+4x^4-12x^3+4x^2+x+1\geq 0$? Question
How do we prove the following for all $x \in \mathbb{R}$ :
$$x^6+x^5+4x^4-12x^3+4x^2+x+1\geq 0 $$
My Progress
We can factorise the left hand side of the desired inequality as follows:
$$x^6+x^5+4x^4-12x^3+4x^2+x+1=(x-1)^2(x^4+3x^3+9x^2+3x+1)$$
However, after this I was unable to make any further progress in deducing the desired inequality.
I appreciate your help
 A: $$
x^4+3x^3+9x^2+3x+1 = x^2\left(x+\frac 32\right)^2 + \frac{27}{4}\left(x+\frac 2 9\right)^2 + \frac 23
$$
is strictly positive for all real $x$.
How did I come up with that? I started by completing the square in
$$
x^4+3x^3+9x^2 = x^2(x^2+3x+9) = x^2\left( (x+\frac 32)^2 + \frac{27}{4}\right)
$$
so that
$$
x^4+3x^3+9x^2 +3x+1= x^2\left( x+\frac 32 \right)^2 +  \frac{27}{4}\left( x^2 +\frac 4 9 x + \frac{4}{27}\right)
$$
and then completed the square in $x^2 +\frac 4 9 x + \frac{4}{27}$.
A: Let us begin with noticing that
\begin{align*}
f(x) & = x^{6} + x^{5} + 4x^{4} - 12x^{3} + 4x^{2} + x + 1\\\\
& = x^{3}\left[\left(x^{3} + \frac{1}{x^{3}}\right) + \left(x^{2} + \frac{1}{x^{2}}\right) + 4\left(x + \frac{1}{x}\right) - 12\right] \tag{$x\neq 0$}
\end{align*}
Hence, if we make the change of variable $u = x + 1/x$, we get that:
\begin{align*}
\left(x^{3} + \frac{1}{x^{3}}\right) + \left(x^{2} + \frac{1}{x^{2}}\right) + 4\left(x + \frac{1}{x}\right) - 12 & = u(u^{2} - 3) + u^{2} - 2 + 4u -12\\\\
& = u^{3} + u^{2} + u - 14\\\\
& = (u^{3} - 2u^{2}) + (3u^{2} - 6u) + (7u - 14)\\\\
& = u^{2}(u - 2) + 3u(u - 2) + 7(u - 2)\\\\
& = (u^{2} + 3u + 7)(u - 2)
\end{align*}
If $x = 0$, the proposed relation clearly holds. If $x > 0$, then $u\geq 2$. Finally, if $x < 0$, then $u\leq -2$.
Gathering all these results, one concludes the validity of the proposed claim.
Hopefully this contributes!
A: $$x^4+3x^3+9x^2+3x+1=\left(x^2+\frac32x+1\right)^2+\frac{19}4x^2$$ is always positive.
There is a general theorem that if a real polynomial is always non-negative, it can be written as a sum of squares.
A: Your quartic polynomial is called self-reciprocal or palindromic, in mathematics.
Let,
$$
\begin{align}P(x):=&x^4+3x^3+9x^2+3x+1\end{align}
$$
The case $x=0$ is trivial. Therefore, we can divide all terms of the polynomial $P(x)$ by $x^2\thinspace : (x\neq 0\thinspace)$
$$
\begin{align}\frac {P(x)}{x^2}=&\left(x^2+\frac {1}{x^2}\right)+3\left(x+\frac 1x\right)+9\end{align}
$$
Then, using the standard substitution $x+\frac 1x=u$, you have:
$$
\begin{align}&\frac {P(x)}{x^2}=\underbrace{ u^2+3u+7}_{\Delta_u<0}\thinspace >0\\
\implies &P(x)>0\thinspace , \forall x\neq 0\end{align}
$$
This means: $\thinspace P(x)>0,\thinspace \forall x\in\mathbb {R}\thinspace.$
A: By using the symmetry in quartic,
\begin{array}
.x^4+3x^3+9x^2+3x+1&=x^2(x^2+3x+\frac{9}{2}+\frac{9}{2}+3x^{-1}+x^{-2})\\
&=x^2\large\left((x+\frac{3}{2})^2+(x^{-1}+\frac{3}{2})^2+\frac{9}{2}\large\right)\geq0
\end{array}
and at $x=0$ it is $1\geq 0$.
A: Using specific properties of this polynomial one can get short proofs, like in all the many answers.
This answer will produce a proof that has more computation, but as a reward it works for any other polynomial with real coefficients and one variable. The algorithm below is not the most efficient either (for a computer), but it is easy to explain and to learn for students and to carry out by hand classroom-type exercises.
Input: The input is an arbitrary non-zero univariate polynomial $f\in\mathbb{R}[x]$.
Output: The algorithm decides of $f(x)\geq0$ for all $x\in\mathbb{R}$ (assuming that we are able to decide the sign of the coefficients and the result of arithmetic operations on the coefficients). In other words, the algorithm decides if $f$ is positive semi-definite, which is the name that is given to this property. It can also decide if $f(x)>0$ for all $x\in\mathbb{R}$.
Step 1: Compute the square-free factorization of $f$. This is a factorization of the form
$$f(x)=a_1(x)a_2(x)^2a_3(x)^3...a_n(x)^n$$
in which all $a_k$ are square-free (their roots have multiplicity $1$) and are pairwise relatively prime ($\gcd(a_i(x),a_j(x))=1$, for $1\leq i< j\leq n$). This factorization can be carried out by Yung's algorithm. Note how all we need to do are arithmetic operations between polynomials and computing $\gcd$, which in turn can be computed by the Euclidean algorithm.
Note that $f$ is positive semi-definite if and only if all $a_k$ have no real roots, for $k=1,3,5,...$ (the ones with odd exponent and subscript).
Step 2: Use Sturm's theorem to determine if each $a_1,a_3,a_5,...$ have no real roots. Note how the application of Sturm's theorem consists of computing a Sturm sequence and looking at the signs of their leading coefficients. In the linked article there is one Sturm sequence that gets computed doing a few polynomial divisions.
In your example:
The result of Step 1 is precisely the factorization that you obtained.
$$
\begin{align}
a_1(x)&=x^4+3x^3+9x^2+3x+1\\
a_2(x)&=x-1
\end{align}
$$
So, Step 2. only needs to be carried out for $a_1(x)$, in your example.
The Sturm sequence in the Wikipedia page for your $a_1(x)$ would be
$$
\begin{align}
x^4+3x^3+9^2+3x+1\\
4x^3+9x^2+18x+3\\
-\frac{45}{16}x^2+\frac{9}{8}x-\frac{7}{16}\\
-\frac{4864}{225}x-\frac{304}{225}\\
\frac{2125}{4096}
\end{align}
$$
Look at the statement of Sturm theorem in the link. The specific values of the coefficients are not important, but only their signs and the degrees of the leading terms. These tell you that $a_1(x)$ has no real roots and that $a_1(x)$ is always positive.
A: Let $P(x) = x^4 + 3x^3 + 9x^2 + 3x + 1$
$P(x≥0) \;≥\; 1 \;>\; 0\;$
$P(y=-x) \;=\; y^4 - 3y^3 + 9y^2 - 3y + 1 = (y-1)^4 + (y^3+3y^2+y)$
$P(x<0 \;⇒\;y>0) \;>\; 0 \quad\quad $ // RHS first term ≥ 0, second term > 0
$\forall x\in\mathbb {R} \;⇒\; P(x)>0$
