Show that $x^{12}-x^9+x^4-x+1\geq0$ for all $x$

Question

Show that $x^{12}-x^9+x^4-x+1\geq0$ for all $x$.

When $x\leq0$ , this is easy.
When $x\geq1$, then also this is easy.
I need help with the case when $0\leq x\leq 1$

Answer 1

$$x^{12}-x^9+x^4-x+1 \geq x^{12}-x^9+x^4 \geq x^4-x^9\geq 0,$$ Where the first inequality uses that $x$ is less than $1$ and the second that $x$ is positive. The last inequality just uses that when you exponentiate by a number greater than $1$, it decreases in size for small $x$.

Answer 2

$x^{12} - x^9 +x^4 - x + 1 \ge 0$;

Rewritten:

$x^{12} + (x^4 - x^9) + (1 - x) \ge 0$.

For $0 \le x \le 1$:

We have $x^{12} \ge 0$; $(x^4 -x^9) \ge 0$ and $(1-x) \ge 0$.

The sum of the above terms is $ \ge 0$.