Minimum distance to points in plane Someone told me that the the following problem is elementary. Given three points $a=(-5,0)$, $b=(0,5)$ and $c=(5,0)$ in $\mathbb R^2$ with Euclidean norm: 
$$\mbox{minimize}\;\; \; f(x)=\|x-a\| + \|x-b\|-2\|x-c\|,\;\; x\in \mathbb R^2.$$
I will highly appreciate for your answers. Thanks in advance. 
 A: You seem to be serious about solving this problem. What I would try is this: Choose any $r$ such that
$$
            \|a-b\| \le r < \infty.
$$
Consider the ellipse
$$
             \|x-a\|+\|x-b\|=r,
$$
for which you can get an equation by the standard trick of expanding:
$$
                     \|x-a\|^{2}=r^{2}-2\|x-b\|+\|x-b\|^{2} \\
                         \|x-a\|^{2}-\|x-b\|^{2}-r^{2}=-2\|x-b\| \\
                (2(b\cdot x)-2(a\cdot x)+\|a\|^{2}-\|b\|^{2}-r^{2})^{2}=4\|x-b\|^{2}.
$$
("$\cdot$" represents vector dot product in the last expression.)
Now, minimize $f$ on this ellipse. This comes down to considering $f(x)=r-2\|x-c\|$ for $x$ on the ellipse. Finally, look at the minimum value of $f$ as a function of $r$, and see what you can do with that.
Note: Try finding the location of the minimum for $f(x)=r-2\|x-c\|$ on the ellipse by finding the location of the maximum for $\|x-c\|^{2}=\|x\|^{2}-2(x\cdot x)+\|c\|^{2}$ on the ellipse, which can be done by first replacing $\|x\|^{2}=x_{1}^{2}+x_{2}^{2}$ with terms from the equation for the ellipse.
No guarantee that this will lead to enough reduction in complexity, but I don't see any other reasonable suggestions. It might be worthwhile to first rotate the configuration until $a$ and $b$ are on one horizontal or on one vertical line.
