Problem 3

Determine the minimum of

a^2+b^2

over real

a,b

for which

x^4+ax^3+bx^2+ax+1=0

has at least one real solution.

Step 2 of 5: Translate the root condition

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In plain words

A real value of $x+1/x$ lies outside the interval $(-2,2)$ .

y_\pm=\frac{-a\pm\sqrt{a^2+8-4b}}2,\quad \max(|y_+|,|y_-|)\ge2

Detailed analysis

The quadratic in $y$ has real roots because the original polynomial has a real root. For a real $x\ne0$ , $y=x+1/x$ satisfies $|y|\ge2$ , so at least one of the two displayed roots must have absolute value at least $2$ .