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 

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.
