Finding the shortest distance between two Parabolas Recently, a problem asked me to find the minimum distance between the parabolas $y=x^2$ and $y=-x^2-16x-65$.
I proceeded with the problem as thus. 
Let $P(a,a^2), Q(b, -b^2-16b-65), a-b=x$.
$\therefore PQ^2=x^2+(2a^2+2ax+16a+x^2+16x+65)^2$.
$PQ^2=x^2+(2(a+\frac{x+8}{2})^2+\frac{(x+8)^2+2}{2})^2 \ge (x^2+(\frac{(x+8)^2+2}{2})^2)(1+\frac{1}{4}) \times \frac{4}{5}$
Applying Cauchy gives us that 
$PQ^2 \ge (\frac{1}{4}x^2+3x+\frac{33}{2})^2 \times \frac{4}{5} \ge (\frac{15}{2})^2 \times \frac{4}{5}=75$
This implies that the answer is $\sqrt{75}$. 
However, it took me a long time to find the values for Cauchy, and the calculations proved tedious. 
What are other approaches to this problem?
EDIT: $(\frac{15}{2})^2 \times \frac{4}{5} \neq 75$, it`s $45$ actually! 

You could define a new function as the distance between a point on one parabola and a point on the other. 
We know a generic point for the first parabola is given by $(x,x^2)$ and a generic point for the second parabola is given by $(y,-y^2-16y-65)$
We want to minimize
$$min{\sqrt{(y-x)^2+(-y^2-16y-65-x^2)^2}}$$
Note that this is the same as minimizing
$$min{(y-x)^2+(-y^2-16y-65-x^2)^2}$$
This is another approach.
