Romanian Maths Olympiad.

The question should state that the constants $a,b$ are real. If $\alpha$ is a root of the polynomial, so is $\alpha^{-1}$ .

Suppose that $\alpha \neq \pm1$ is a real zero of the polynomial. Then $\alpha^{-1}$ is a distinct zero, and hence $x^2 + cx + 1$ is a factor of the polynomial, where $c = -\alpha - \alpha^{-1}$ , so that $|c| > 2$ .

If $1$ is a repeated root of the polynomial, then $x^2 - 2x + 1$ is a factor of the polynomial. If $-1$ is a repeated root of the polynomial, then $x^2 + 2x + 1$ is a factor of the polynomial.

If no $\alpha \neq \pm1$ is a root of the polynomial, and neither $1$ nor $-1$ is a repeated root of the polynomial, then (since the polynomial must have two real roots) both $1$ and $-1$ must be non-repeated roots of the polynomial, and so $x^2 - 1$ is a factor. But the only way that $x^2 - 1$ can divide a polynomial of the form $x^4 + ax^3 + bx^2 + ax +1$ is when $x^4 - 2x^2 + 1 \; = \; (x^2 -1)(x^2 - 1)$ in which case both $1$ and $-1$ are repeated roots.

Thus we deduce that $x^2 + cx + 1$ is a factor of the polynomial for some real $c$ with $|c| \ge 2$ . But then $x^4 + ax^3 + bx^2 + ax + 1 \; = \; (x^2 + cx + 1)(x^2 + dx + 1)$ for some real $d$ , so that $a = c+d$ and $b = cd + 2$ , and hence $a^2 + b^2 \; = \; (c+d)^2 + (cd + 2)^2$ For a fixed value of $c$ , the above expression is minimised when $d = -\tfrac{3c}{c^2+1}$ , and the minimum value is $\tfrac{(c^2 - 2)^2}{c^2 + 1}$ . Thus the minimum value, for $|c| \ge 2$ , occurs when $|c|=2$ , and the minimum value is $\boxed{\tfrac45}$ .

Firstly note that $a^2+b^2>0$ as since if $a^2+b^2=0$ was true then $a=b=0$ and the equation $x^4+1=0$ wouldn't have a real solution.

Divide by $x^2$ throughout and obtain,

$\displaystyle (x+\tfrac 1x)^2+a(x+\tfrac 1x)+(b-2)=0$ and Set $x+\tfrac 1x\equiv y$ so that ,

$\displaystyle y^2+ay+(b-2)=0$

It' solution is given by $\displaystyle y=\frac{-a\pm\sqrt{a^2-4b+8}}{2}=\zeta$ (Say)

Substitute back $x$ and we find the equation to be $x^2-\zeta x+1=0$

For it's roots to be real we must have , $\displaystyle \zeta^2\ge 4\implies |-a\pm\sqrt{a^2-4b+8}|\ge 4$

We accept the case $\displaystyle -a\pm\sqrt{a^2-4b+8}>4$

Simplifying the inequality would produce $2a+b\le -2$

Now We consider the Lagrangian $\displaystyle L_\lambda(a,b)=a^2+b^2+\lambda(-2-2a-b)$

Equating the partial derivatives to zero gives, $\displaystyle \begin{aligned} \frac{\partial{L}}{\partial{a}}=0 \implies a=\lambda \\ \frac{\partial{L}}{\partial{b}}=0\implies b=\frac{\lambda}{2}\end{aligned}$

Now substituting $a,b$ in the constraint we have, $\lambda=-0.8$ . To check if it's a minimum we consider it's Hessian

$\displaystyle \begin{pmatrix} \frac{18}{5} & 0\\0&\frac{14}{5}\end{pmatrix}$ which is positive definite and thus it's a minimum.

So, $a^2+b^2\ge 0.8$

If we have taken the case $-a\pm\sqrt{a^2-4b+8}<-4$ and proceeded in the same way then we would have got the minimum as $(0,0)$ which is not possible.