Finding the minimum value of a 6th degree polynomial algebraically

Question

Is it possible to answer this question using methods of basic algebra? Find the least value of the expression $a^6 + a^4 - a^3 - a + 1$ for real value of $a$. This question is from the 2013 Philippine Mathematical Olympiad, Oral Stage.

Answer 1

The minimium of $f(a)=a^6+a^4-a^3-a+1$ is at a point where $f'(a)=0$, i.e. $6a^5+4a^3-3a^2-1=0$. Clearly $f'(a)\le 1<0$ for $a\le 0$, i.e. $f'$ has only positive roots. If $0<a_1\le a_2$ are two roots of $f'$, then between them there is a root of $$f''(a)=30a^4+12a^2-6a=6a(5a^3+2a-1)$$ and hence of $$\tag1 5a^3+2a-1.$$ The derivative of this is $15a^2+2\ge 2>0$, hence there is at most one simple positive root of $f''$, hence at most two positive roots of $f'$. From $f'(0)<0$ and $f'(a)\to+\infty$ as $a\to+\infty$, we conclude that the number of positive roots of $f'$ must be odd, hence there is exactly one such root. From $f'(\frac12\sqrt 2)=\frac74\sqrt 2-\frac52<0$, we see that the root of $f'$ is $>\frac12\sqrt 2$, hence the minimum of $f$ is definitely less than $f(\frac12\sqrt2)=\frac{11}8-\frac34\sqrt 2\approx 0.3143398$, but I doubt that the exact value $f(0.7097999841253\ldots)\approx 0.31430590578\ldots$ can be expressed with basic algebra - after all the polynomial $6a^5+4a^3-3a^2-1$ is irreducible over the raionals.