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 4 of 5: Optimize over

a

and then

|y|\ge2

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In plain words

For each possible root value $y$ , first choose the best coefficient $a$ ; the remaining one-variable expression is smallest at the boundary $|y|=2$ .

a=\frac{y(2-y^2)}{1+y^2},\quad b=\frac{2-y^2}{1+y^2},\quad a^2+b^2=\frac{(y^2-2)^2}{1+y^2}\ge\frac45

Detailed analysis

For fixed $y$ , minimizing $a^2+(2-y^2-ay)^2$ gives $a=y(2-y^2)/(1+y^2)$ and $b=(2-y^2)/(1+y^2)$ . Substitution gives the displayed value. Put $t=y^2\ge4$ ; the function $(t-2)^2/(t+1)$ is increasing for $t\ge4$ , so it is at least $(4-2)^2/5=4/5$ .