parmenides51 wrote: Let $a$ and $b$ be real numbers. It is known that the graph of the parabola $y =ax^2 +b$ cuts the graph of the curve $y = x+1/x$ in exactly three points. Prove that $3ab < 1$. $ax^2+b=x+\frac 1x$ has exactly three distinct roots $\iff$ $ax^3-x^2+bx-1=0$ has exactly three distinct roots $\implies$ (derivative) $3ax^2-2x+b$ has exactly two distinct roots $\implies$ (determinant) $1-3ab>0$ Q.E.D.