Question: Find the minimum possible value of $\displaystyle a^{2}+b^{2}$such that the quartic polynomial $\displaystyle x^{4}+ax^{3}+bx^{2}+ax+1$ has at least 1 real root.

Attempt: Since the equation is palindromic, the standard technique is:
$\displaystyle x^{4}+ax^{3}+bx^{2}+ax+1\Rightarrow x^{2}+ax+b+\frac{a}{x}+\frac{1}{x^{2}}$
$\displaystyle Let X=x+\frac{1}{x}\therefore X^{2}+pX+q=0$

Since $\displaystyle \frac{f(x))}{x^{4}}=f(\frac{1}{x})=0$the real roots must be 1 or -1 correct? But when I solve simultaneous equations using 1 and -1 as roots I get $\displaystyle a^{2}+b^{2}=4$which is too large. The glitch I ran into was excluding the case that at the minimal value of $\displaystyle a^2+b^2$there might be two real roots of multiplicity 2. That made things a little ugly.

I also tried this approach: Suppose r is a simple multiplicity 1 root and a^2+b^2 is at it's minimal value. The graph of the polynomial near the root the just looks like a line crossing the x-axis, right? That means if I move the graph just a little by varying a and b, there will still be a real root because the graph will still cross the x-axis, just in a slightly different spot. That in turn means I can make a or b just a little smaller so I decrease $\displaystyle a^2+b^2$and still have a root. So that would contradict $\displaystyle a^2+b^2$being the smallest value where you still have a real root.
That means at the minimal value, all of the real roots must have multiplicity 2 or greater but I think I am stuck here...

Bump!