Minimizing $l(x,y)$

Let us, for avoiding confusion later, write $l$ as $l(a,b)=(a-b)^2+(a^2-3b+5)^2 \ \ \forall \ a,b \in \mathbb{R}$

Now, this looks similar to the distance formula in coordinate geometry, doesn't it? If this strikes , then the problem becomes easy.

We can think of $l(a,b)$ as the square of the distance between the points $(a,a^2+5)$ and $(b,3b)$ .

The first set of points is the upward parabola $y=x^2+5$ and the second set of points is the straight line $y=3x$ .

$\therefore$ we need to find the shortest distance between the curves $y=3x$ and $y=x^2+5$ .

This occurs along the common normal, as can be seen from a rough sketch of the graphs.

Or, this can be put the other way round, and we can say that the tangent at the point on the parabola which is nearest to the line is parallel to the line.

$\therefore \ 2x=3 \ \implies \ x=\dfrac{3}{2}$ .

The point on the parabola nearest to the line is therefore $\left(\dfrac{3}{2}, \dfrac{29}{4}\right)$

Hence, $\begin{aligned} \min \{l(a,b)\}=\min \{l(x,y)\} &=\left|\dfrac{3 \cdot \dfrac{3}{2}-\dfrac{29}{4}}{\sqrt {3^2+1^2}}\right|^2\\ &= \left|-\dfrac{11}{4\sqrt 10}\right|^2 \\ &=\boxed{\dfrac{121}{160}} \end{aligned}$