<s>[INST] Condition for quartic polynomial coefficients given at least one real root 
Find the minimum possible value of $a^2+b^2$ where $a$, $b$ are two real numbers such that the polynomial
  $$x^4+ax^3+bx^2+ax+1,$$
  has at least one real root.

My attempt: Let p be a real root. 
Therefore $p^4 + a(p^3) + b(p^2) + ap + 1 = 0$. 
Divide both sides by $p^2$, noting that $p$ cannot be zero since $P(0) = 1$, we get 
$$p^2 + ap + b + a/p + 1/(p^2) = 0.$$
I rearranged this into a quadratic; i.e. 
$$(p+1/p)^2 + a(p+1/p) + (b-2) = 0.$$
If $p$ is real, $p+1/p$ is real, so discriminant is non-negative.
setting discriminant $\geq0$, I get:
$$a^2 - 4(b-2)\geq0,$$
$$a^2 \geq 4b-8,$$
$$a^2 + b^2 \geq b^2 + 4b - 8 = (b+2)^2 - 12 \geq -12$$
But this is useless because it is obvious $a^2 + b^2 \geq 0$. 
I believe this is because I also need to use the fact that $|p+1/p| \geq 2$, however I am unsure how to use this inequality with the discriminant. 
Thanks in advance
 [/INST] Replacing $a$ with $-a$ only changes the signs of the zeros, so w.l.o.g. we can assume that $a\ge0$. The quadratic with the unknown $p+1/p$ (nice trick, BTW!) that you derived gives
$$
p+\frac1p=\frac{-a\pm\sqrt{a^2-4(b-2)}}2.
$$
Given the assumption $a\ge0$ we see that of these two alternatives the solution with a minus sign gives the larger value to $|p+1/p|$. Therefore the condition $|p+1/p|\ge2$, now rewritten as $p+1/p\le-2$, gives us another constraint
$$
-a-\sqrt{a^2-4(b-2)}\le-4.
$$
This is equivalent to
$$
4-a\le\sqrt{a^2-4(b-2)}.\qquad(*)
$$
We shall see that there are solutions with $a^2+b^2<16$, so w.l.o.g. we can further assume that $a<4$, so $4-a>0$ and we can square both sides of $(*)$ arriving at
$$
16-8a+a^2\le a^2-4b+8\implies a\ge (b+2)/2.
$$
Again, later developments will reveal that $b+2$ must be positive at the sought minimium, so to minimize $a^2$ we must have equality here, i.e. $a=(b+2)/2$. Thus we really want to minimize
$$
a^2+b^2=\frac14[(b+2)^2+4b^2]=\frac14[5b^2+4b+4].
$$
It is trivial to show that this has a minimum at $b=-2/5$. The corresponding
$a=(b+2)/2=4/5$, and at this point we have
$$
a^2+b^2=\frac{4^2+2^2}{25}=\frac45.
$$
If either the assumption $4-a\ge0$ or the assumption $b+2\ge0$ (that I made while looking for this candidate point) were invalid, then clearly $a^2+b^2$ would have a larger value, so we can dismiss those possibilities.
As a final check we see that when $a=4/5, b=-2/5$ your polynomial has a double root at $x=-1$, not unexpectedly matching with $p+1/p=-2$.
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