How find this $3\sqrt{x^2+y^2}+5\sqrt{(x-1)^2+(y-1)^2}+\sqrt{5}(\sqrt{(x-1)^2+y^2}+\sqrt{x^2+(y-1)^2})$ find this follow minimum
$$3\sqrt{x^2+y^2}+5\sqrt{(x-1)^2+(y-1)^2}+\sqrt{5}\left(\sqrt{(x-1)^2+y^2}+\sqrt{x^2+(y-1)^2}\right)$$
I guess  This minimum is $6\sqrt{2}$
But I can't prove,Thank you 
 A: If $v_1 = (0,0), v_2 = (1,1), v_3 = (0,1)$, and $v_4 = (1,0)$ and $p = (x,y)$, then you are trying to minimize $$3|p - v_1| + 5|p - v_2| + \sqrt{5}|p - v_3| + \sqrt{5}|p - v_4|$$Note that if $p$ is on the line $y = x$, moving it perpendicularly away from the line will only increase $|p - v_1|$ and $|p - v_2|$, and it is not too hard to show it also increases $|p - v_3| + |p - v_4|$. So the minimum has to occur on the line $y = x$. So letting $p = (t,t)$ your problem becomes to minimize
$$3\sqrt{2}t + 5\sqrt{2}(1 - t) + 2\sqrt{5}\sqrt{2t^2 - 2t + 1}$$
This can be minimized through calculus... maybe there's a slick geometric way too.
A: Too long for a comment, I wonder if the problem can be solved geometrically.
Let $A,B, C, D$ be the points of coordinates $(0,0), (1,1), (1,0)$ and $(0,1)$ and let $P$ be the point in plane of coordinates $(x,y)$.
The problem asks you to find the minimum of 
$$3PA+5PB+\sqrt{5}(PC+PD) (*)$$
The issue now is that we cannot split the problem in two halves: $\sqrt{5}(PC+PD) \geq \sqrt{5} CD$ with equality if and only if $P \in (C,D)$, but $3PA+5PB \geq 3AB$ with equality if and only if $P=B$. 
But one might be able to find the min of $(*)$ geometrically...Or maybe physically (put four objects of weights $3,5,\sqrt{5}, \sqrt{5}$ at $A,B,C,D$ then $P$ must have some physical meaning).
