f(x,y)=(x^2+y^2)^(1/2)+((a-x)^2+(b-y)^2)^(1/2)+((c-x)^2+y^2)^(1/2)
whats the minimum of this function in function of a,b,c parameters?
tried simple methods but none of them worked
My recommendation would be to take partial derivatives with respect to $x$ and $y$:
$f_x = \dfrac{x}{\sqrt{x^2+y^2}} + \dfrac{x-a}{\sqrt{(a-x)^2+(b-y)^2}} + \dfrac{x-c}{\sqrt{(c-x)^2+y^2}}$
$f_y = \dfrac{y}{\sqrt{x^2+y^2}} + \dfrac{y-b}{\sqrt{(a-x)^2 + (b-y)^2}} + \dfrac{y}{\sqrt{(c-x)^2+y^2}}$
Set each equal to zero and solve. This will give you the minimums along each axis. This does not guarantee that these will equate to a local minima for the function in general. I do not remember what criteria there was for determining if minimizing the partial derivatives would suffice for minimizing the whole function. It has been years since I studied multivariate analysis, so I only dimly recall that this method does not always produce actual minima.
What "simple methods" did you try? The usual is to set the partial derivatives equal to 0.
The derivative of f with respect to x is $\displaystyle x(x^2+ y^2)^{-1/2}- (a- x)((a- x)^2+ (b- y)^2)^{-1/2}- (c- x)((c- x)^2+y^2)^{-1/2}= 0$
and the derivative with respect to y is $\displaystyle b(x^2+ y^2)^{-1/2}- (a- y)((a- x)^2+ (b- y)^2)^{-1/2}+ y((c- x)^2+y^2)^{-1/2}= 0$
Actually, we might be able to do a little better. The first term is minimized at $(0,0)$. The second term is minimized at $(a,b)$. The third term is minimized at $(c,0)$. My guess would be the whole function is minimized at the center of the triangle created by the three points. My guess is that you can prove this is a minimum using the triangle inequality a bunch of times?
So, I would venture the guess that the minimum occurs at: $\left(\dfrac{a+c}{3},\dfrac{b}{3}\right)$
I've tried the partial differentiation method but i found these equations too hard to solve, I wonder if theres any method which would give us the local minimum(s) of this function.
Lol, I gave you a general method that does not require presumptions. You just don't want to do any work to prove it. Like I said, I bet it is easily proven with a few applications of the Triangle Inequality that the point I found is the location of the absolute minimum. That is as general a method as there is. I just used a little geometric intuition.
In geometry, the Fermat point of a triangle, also called the Torricelli point or Fermat–Torricelli point, is a point such that the total distance from the three vertices of the triangle to the point is the minimum possible.[1] It is so named because this problem is first raised by Fermat in a private letter to Evangelista Torricelli, who solved it.
From "Wikipedia"
You clearly need more help than can be given through a math help forum. If you don't understand how my answer related to the question asked, then you need a LOT more help than we can provide.
$\sqrt{x^2+y^2}$ is the Euclidean distance from the point $(0,0)$ to the point $(x,y)$.
$\sqrt{(a-x)^2+(b-y)^2}$ is the Euclidean distance from the point $(a,b)$ to the point $(x,y)$.
$\sqrt{(c-x)^2+y^2}$ is the Euclidean distance from the point $(c,0)$ to the point $(x,y)$.
Three coplanar points create a triangle. If they are collinear, a line. Regardless, as HallsofIvy pointed out, Fermat and Torricelli proved that the point you are looking for is the exact point that gives the shortest total distance to the three vertices of the triangle. You are just too lazy to look it up. We literally did your homework for you, and all you can do is complain.