69th Belarusian Mathematical Olympiad

Since the polynomial $f(x)=x^3+a{x}^2+2bx-1$ has three different roots, the equation $$x^3+a{x}^2+2bx-1=0(1)$$ has three different solutions, denote them by $x_1$, $x_2$ and $x_3$. Vieta's formulas implies the equalities $x_1+x_2+x_3=-a$, $x_1{x}_2+x_1{x}_3+x_2{x}_3=2b$ and $x_1{x}_2{x}_3=1$. Hence $$x_1^2+x_2^2+x_3^2=(x_1+x_2+x_3{)}^2-2(x_1{x}_2+x_1{x}_3+x_2{x}_3)=a^2-4b.$$ From the AM-GM inequality: $$a^2-4b=x_1^2+x_2^2+x_3^2>3\sqrt{x_1^2{x}_2^2{x}_3^2}=3(2)$$ (the inequality is strict, since the solutions of (1) are distinct).

The quadratic polynomial $g(x)$ doesn't have roots, therefore its discriminant is negative, i.e. $4b^2-8a<0$, whence $2a-b^2>0$. Summing this inequality with (2) we obtain $(2a-b^2)+(a^2-4b)>3$, which is equivalent to $(a+1{)}^2-(b+2{)}^2+3>3$. Therefore, $$(a+1+b+2)(a+1-b-2)>0.$$ Since $a$ and $b$ are positive, $a+b+3>0$, so the last inequality implies $a-b-1>0$.