I have the question: to find the minimum value of$\sqrt{x^2+(20-y)^2}+\sqrt{y^2+(21-z)^2}+\sqrt{z^2+(20-w)^2}+\sqrt{w^2+(21-x)^2}$.
My method, by making use of Pythagoras' theorem($a^2+b^2=c^2$), the 2 terms inside the square root must be equal to be minimum. Using this fact, for each square root, I draw 2 lines to represent each term, $a$ and $b$, in the square root, and a diagonal to be the $c$. Then I connect the diagonals, which is my desired solution. The $x$ cancels out with the $21-x$, and so on... I get $\sqrt{42^2+40^4}=58$.
Now, I have made use of geometry. Is there another way to solve this?
By Minkowsky we obtain: $$\sqrt{x^2+(20-y)^2}+\sqrt{y^2+(21-z)^2}+\sqrt{z^2+(20-w)^2}+\sqrt{w^2+(21-x)^2}\geq$$ $$\geq\sqrt{(x+21-z+z+21-x)^2+(20-y+y+20-w+w)^2}=\sqrt{42^2+40^2}$$