Q:

What is the minimum value of $A$?

Let x, y be real numbers and let  
$A=\sqrt{x^2+4}+\sqrt{(y-x)^2+9}+\sqrt{(9-y)^2+49}$.
How do I simply find the minimum value of A? I know the answer is 15 but I need to express the solution in simple steps. I suspect it has no global minima but local. 

A:

By Minkowski
$$\sqrt{x^2+4}+\sqrt{(y-x)^2+9}+\sqrt{(9-y)^2+49}\geq$$
$$\geq\sqrt{(x+y-x+9-y)^2+(2+3+7)^2}=15$$
The equality occurs for $$(x,y-x,9-y)||(2,3,7),$$
which gives $(x,y)=\left(\frac{3}{2},\frac{15}{4}\right)$.
Id est, the answer is $15$.

