Question about finding minima using Minkowsky's inquality 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?

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).
