Prove the polynomial $x^4+4 x^3+4 x^2-4 x+3$ is positive Given the following polynomial
$$
x^4+4 x^3+4 x^2-4 x+3
$$
I know it is positive, because I looked at the graphics

and I found with the help of Mathematica that the following form
$$
(x + a)^2 (x + b)^2 + c^2(x + d)^2 + e^2
$$
can represent the polynomial with the following values for the constants
$$
\left(x-\frac{1}{2}\right)^2
   \left(x+\frac{5}{2}\right)^2+\frac{5}{2} \left(x+\frac{1}{5}\right)^2+\frac{107}{80}
$$
I suppose there are simpler ways to prove that the polynomial is positive, perhaps by using some inequalities.
Please, advice.
 A: Here's a slightly more motivated way to find a sum-of-squares representation.
We should want the first term to be the square of a quadratic with leading coefficient $1$, and to cancel out the $x^3$ term we need it to be $(x^2+2x+a)^2$ for some real $a$. This will leave us to show that
$$-2ax^2-(4a+4)x+(3-a^2)\geq 0;$$
the simplest case of this is to choose $a=-1$ (to remove the $x$ term entirely), which gives us that
$$x^4+4x^3+4x^2-4x+3=(x^2+2x-1)^2+2x^2+2>0.$$
A: if $x\ge 0$ by AM-GM $$x^4+3=x^4+1+1+1\ge 4x$$  and the inequality $$\color{blue}{x^4+3-4x}+4x^3+4x^2>0$$ is obvious as equality of am-gm and $x=0$ is simultaneously is not achieved.
If $x\le 0$ then replace $x$ by $-t $ and we have to prove $$t^4-4t^3+4t^2+4t+3>0$$which is true as $$t^4+4t^2\ge 4t^3$$ by AM-GM and as  $t\ge 0$ and the rest of terms are postive (note :again equality of both AM-GM and t=0 is not achieved)
A: Just another solution (similar to Albus Dumbledore's).
$$x^4+4x^3+4x^2-4x+3=x^2(x+2)^2-4x+3 = x^4+4x^3+(2x-1)^2 +2$$
For $x<0$ we have $-4x+3>0$ and therefore $x^2(x+2)^2-4x+3>0$.
For $x>0$ it follows from $x^3>0$ that $x^4+4x^3+(2x-1)^2 +2>0$.
A: Let the polynomial be denoted by $f(x)$.
$f'(x)=4(x^3+3x^2+2x-1)$ has only one real root $\approx 0.3247$. At that point $f''(x)=4(3x^2+6x+2)>0$. Thus $f(x)$ has minimum value at $\approx 0.3247$ and $f(0.3247)\approx 2.271$.
$\therefore f(x)>0 \;\forall x\in \mathbb R$.
