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

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.
