What is the minimum value of $A$?

Question

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.

Answer 1

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 $15$.