problem: The smallest distance between the origin and a point on the graph of $y=\frac{1}{\sqrt{2}}\left(x^2-3\right)$ can be expressed as $\sqrt{a}/b$, where $a$ and $b$ are positive integers such that $a$ is not divisible by the square of any integer greater than one.  Find $a+b$.
solution: By the distance formula, we are trying to minimize $\sqrt{x^2+y^2}=\sqrt{x^2+(1/2)(x^4-6x^2+9)}$. In general, minimization problems like this require calculus, but one optimization method that sometimes works is to try to complete the square.  Pulling out a factor of $1/2$ from under the radical, we have \begin{align*}
  \frac{1}{\sqrt{2}}\sqrt{2x^2+x^4-6x^2+9}&=\frac{1}{\sqrt{2}}\sqrt{(x^4-4x^2+4)+5} \\
  &= \frac{1}{\sqrt{2}}\sqrt{(x^2-2)^2+5}.
  \end{align*}This last expression is minimized when the square equals $0$, i.e. when $x=\sqrt{2}$. Then the distance is $\sqrt{5}/\sqrt{2}=\sqrt{10}/2$.  Hence the desired answer is $\boxed{12}$.