How to show this inequality $ a^{14}-a^{13}+a^8-a^5+a^2-a+1>0$

Question

Given the real $ a $. Prove that $$a^{14}-a^{13}+a^8-a^5+a^2-a+1>0$$

I tried to factor it as

$$(a-1)\Bigl(a^{13}+a^5(a^2+a+1)+a\Bigr)+1$$

I think it should be written as a sum of squares. Any idea will be appreciated.

Answer 1

Notice that quadratic $x^2-x+1>0$ for all $x$. Now write: $$ \underbrace{(a-1)(a^{13}-1)}_{\geq 0} + a^2(\underbrace{a^6-a^3+1}_{> 0})>0$$

Answer 2

Notice it a bunch of $+ a^{even}$ and $-a^{odd}$ in descending order of powers so.....

Case 1: $a \le 0$ then $a^{even} \ge 0$ and $-a^{odd} \ge 0$ so $a^{14} - a^{13}+a^8-a^5+a^2 -a + 1=|a|^{14}+|a|^{13}+ |a|^8 +|a|^5 + |a|^2+|a| + 1\ge 1$.

Case 2: $a\ge 1$ then $a^{big} \ge a^{small}$ so $a^{big}-a^{small} >\ge 0$ and $(a^{14}-a^{13}) + (a^8-a^5) + (a^2 -a) + 1 \ge 1$.

Case 3: $0 < a < 1$ then $a^{big} < a^{small}$ so $-a^{big}+a^{small} > 0$ and $a^{14} + (-a^{13}+a^8) + (-a^5+a^2) + (-a + 1) > a^{14} > 0$.

And there aren't any more options.