<s>[INST] How to show this inequality $ a^{14}-a^{13}+a^8-a^5+a^2-a+1>0$ 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.
 [/INST] 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.
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