IMO Hk TST

The minimum value of $a^2 + b^2$ is $\frac{4}{5}$.

WLOG assume $a \ge 0$. (Indeed, by flipping the sign of $a$, we only need to flip the sign of the root $x$.) Clearly, $x \ne 0$. Rewrite the equation as
$$x^2 + a x + b + \frac{a}{x} + \frac{1}{x^2} = 0.$$
This is the same as
$$\left(x + \frac{1}{x}\right)^2 + a\left(x + \frac{1}{x}\right) + b - 2 = 0.$$
Let $y = x + \frac{1}{x}$. Note that $|y| = |x| + \frac{1}{|x|} \ge 2$ by the AM-GM inequality. The equation becomes
$$y^2 + a y + (b - 2) = 0.$$
If $a > 1$, then $a^2 + b^2 > 1$. Thus, we only need to consider the case $a \le 1$. This implies the axis of symmetry $y = -\frac{a}{2}$ lies between $-2$ and $2$.

* If $y \ge 2$, then we have
$$0 = y^2 + a y + (b-2) \ge (2)^2 + a(2) + (b-2) = 2a + b + 2.$$
This yields $b \le -2a - 2 \le -2$, and hence $a^2 + b^2 \ge 4$.

* If $y \le -2$, then we have
$$0 = y^2 + a y + (b-2) \ge (-2)^2 + a(-2) + (b-2) = -2a + b + 2.$$
This yields $b \le 2a - 2 \le 0$, and hence
$$a^2 + b^2 \ge a^2 + (2a - 2)^2 = 5\left(a - \frac{4}{5}\right)^2 + \frac{4}{5} \ge \frac{4}{5}.$$
Equality holds when $a = \frac{4}{5}$ and $b = -\frac{2}{5}$. Correspondingly, this means $y = -2$ and $x = -1$.

To conclude, the minimum value of $a^2 + b^2$ is $\frac{4}{5}$.