Find a point $P$ such that $PA+PB+PC+PD$ has minimum value, where $A(0,0), B(2,0), C(2,1), D(10,1).$
I have assume that point $P$ be $(x,y)$. Then we have:
$PA=\sqrt{x^2+y^2}$ and $PB=\sqrt{(x-2)^2+(y-0)^2}$.
And $PC=\sqrt{(x-2)^2+(y-1)^2}$ and $PD=\sqrt{(x-10)^2+(y-1)^2}$.
Then we have to find minimum value of $\sqrt{x^2+y^2}+\sqrt{(x-2)^2+y^2}+\sqrt{(x-2)^2+(y-1)^2}+\sqrt{(x-10)^2+(y-1)^2}$.
From above it seems that we have to apply triangle inequality. But I did not understand how to apply for $4$ individual parts.
Achille Hui gave a very valuable hint. Let me formulate his idea.
From the triangle inequality, we have that $ \vert AD \vert \leq \vert PA \vert + \vert PD \vert $ and $ \vert BC \vert \leq \vert PB \vert + \vert PC \vert $. Thus, $ \vert PA \vert + \vert PB \vert + \vert PC \vert + \vert PD \vert \geq \vert AD \vert + \vert BC \vert = 1 + \sqrt{101} $. Therefore, $ 1 + \sqrt{101} $ is a lower bound. In fact, the function attains this value at $ P = (2, 0.2) $ which lies at the intersection of the lines $ AD $ and $ BC $.