How do we prove $x^6+3x^3+2x^2+x+1 \geq 0$ Question
How do we prove
$$x^6 + 3x^3+2x^2+x+1\geq0$$
My progress
$$x^6+3x^3+2x^2+x+1=(x+1)^2(x^4-2x^3+3x^2-x+1)$$
I appreciate your interest
 A: Case 1: $x$ is nonnegative.
The last term is greater than or equal to $(x^2-x+1)^2,$ since it is $(x^2-x+1)^2 + x.$ By the Trivial Inequality, both factors are positive or zero, so the product is also positive or zero.
We can also see this directly from the question. It is clear that the $x^6+3x^3+2x^2+x+1 \geq 0,$ since all the summands are nonnegative.
Case 2: $x$ is negative.
We can write this as $(x^2 + 1)^2 + x^2 + (-2x^3-2x).$ Since the third term is positive (since $x$ is negative) and the first two terms are nonnegative, the second factor is positive. Since the first factor is nonnegative, we are done.
Alternatively, we can write the second term as a sum of positive numbers multiplied by squares. It is $(x^2-x)^2 + \frac{7}{4}x^2 + (1-\frac{x}{2})^2$.
A: Again focusing on $x^4-2x^3+3x^2-x+1$, write this as the sum of $f(x) = x^4 - 2x^3 + 3x^2$ and $g(x) = -x + 1$. $f(x) = x^2(x^2 - 2x + 3) = x^2 ((x - 1)^2 + 2)$ which is always nonnegative.
That leaves $g(x)$ which is negative for $x \ge 1$. However, we just need to show that $f(x) > x$ in this domain, which can be shown as $f(x) = (x^2-x)^2 + x^2 \cdot 2 \ge x^2 \cdot 2 \ge 2x > x$. Thus from the first part, $f(x)$ and $g(x)$ are individually nonnegative for all real $x \le 1$, and from the last part, $f(x) + g(x)$ $> x + (1 - x) = 1$ for all real $x > 1$.
Hence proven.
A: $$x^6 + 3x^3+2x^2+x+1= (x + 1)^2 \left( {\left( {x - \tfrac{1}{2}} \right)^4  + \tfrac{1}{2}\left( {x - \tfrac{1}{2}} \right)^2  + x^2  + \tfrac{{13}}{{16}}} \right) \ge 0$$
