Minimum distance between the curves of two inverse functions? What is the minimum distance between a point on the curve y=e^x and a point on the curve y=ln(x)?
What I did: If the curves given were standard one I could find the common normal and solve the equation of the curves and the common normal to find the point of intersection and then use distance formula to find the distance between them. But, sics they are not standard I could not use this method. I could realize that the functions given are inverse of each other and hence they are symmetric about the line y=x.But, how to use this information to solve the problem? 
 A: Consider the orthogonal projection onto the  line $x+y=0$. The distance between orthogonal projections of two points does not exceed the distance between points themselves (for the same reason that a leg of a right triangle is not longer than its hypotenuse: Pythagorean theorem). Therefore, if we can find the distance between projections of two curves, that will be a lower bound on the distance between the curves themselves. With luck and foresight, the lower bound will be sharp, i.e., it will be the actual distance. 
So, what is the projection of $\{(x,y):y=e^x\}$ onto the line $x+y=0$? It's a closed half-line. Its endpoint comes from $(x,e^x)$ such that $e^x-x$ is minimal. Minimizing $e^x-x$ over $x$, we see that the minimum is at $x=0$. The point $(0,1)$ projects to $(-1/2,1/2)$. 
By symmetry, the projection of $\{(x,y):y=\ln x\}$ onto the line $x+y=0$ is a closed half-line with endpoint $(1/{2},-1/2)$. 
The distance between projections is $\sqrt{1^2+1^2} = \sqrt{2}$. Therefore, the distance between curves is at least $\sqrt{2}$. This is the lower bound. 
And it's sharp: the distance between the aforementioned point $(0,1)$ and its counterpart $(1,0)$ is $\sqrt{2}$. 
