Question about finding minima using Minkowsky's inquality

Question

I would like to find the minima of the value $$\sqrt{(x-1)^2+(x-2)^2}+\sqrt{(x-y)^2+(x)^2}+\sqrt{(y-2)^2+(1)^2}$$ Given that $x$ and $y$ are both real numbers. I know that I can rewrite this as $\sqrt{(2-x)^2+(1-x)^2}+\sqrt{(x-y)^2+(x)^2}+\sqrt{(y-2)^2+(1)^2}$ or $\sqrt{(1-x)^2+(2-x)^2}+\sqrt{(x-y)^2+(x)^2}+\sqrt{(y-2)^2+(1)^2}$ But applying Minkowsky's inequality gives value $\sqrt4$ or $\sqrt{10}$, how to deal with this problem?

Answer 1

You need to consider the equality condition for Minkowski's inequality.

Using Minkowski's inequality, we have \begin{align*} &\sqrt{(1-x)^2+(2-x)^2}+\sqrt{(x-y)^2+(x)^2}+\sqrt{(y-2)^2+(1)^2} \\ \ge\,& \sqrt{(1 - x + x - y + y - 2)^2 + (2 - x + x + 1)^2}\\ =\,& \sqrt{10}. \end{align*} Equality holds if $x = 5/4, y = 5/3$ (when $(1 - x, x - y, y - 2)$ and $(2 - x, x, 1)$ are proportional).

On the other hand, using Minkowski's inequality, we have \begin{align*} &\sqrt{(2-x)^2+(1-x)^2}+\sqrt{(x-y)^2+(x)^2}+\sqrt{(y-2)^2+(1)^2}\\ \ge\,&\sqrt{(2 - x + x - y + y - 2)^2 + (1 - x + x + 1)^2}\\ =\,& \sqrt{4}. \end{align*} Equality never holds ($(2 - x, x - y, y - 2)$ and $(1- x, x, 1)$ are never proportional).