<s>[INST] Computing local and global minima on Wolfram Alpha Consider $\sqrt{x^2+y^2}+2\sqrt{x^2+y^2-2x+1}+\sqrt{x^2+y^2-6x-8y+25}$. I need to find global or local minima of this function, but Wolfram Alpha doesn't seem to find one; the answer is that $1 + 2\sqrt{5}$ is its global minimum
Am I doing something wrong? I use this input for the global minimum, and this input for local minima.
 [/INST] Using FindMinimum[Sqrt[x^2+y^2]+2Sqrt[x^2+y^2-2x+1]+Sqrt[x^2+y^2-6x-8y+25],{x,y}] locally, I found the following minimum solutions. The output includes the already mentioned solution $5.47214=1+2\sqrt{5}$ and concretely looks as follows:
$$\{5.47214,\{x\to 1,y\to -7.441589669722822*10^{-10}\}\}$$
Hope it helps you to proceed. The plot is:

And here is the plot of the curve given by $\sqrt{x^2-2 x+y^2+1}+\sqrt{x^2-6 x+y^2-8 y+25}+\sqrt{x^2+y^2}-z=0$:

The partial derivate of $2\sqrt{x^2-2 x+y^2+1}+\sqrt{x^2-6 x+y^2-8 y+25}+\sqrt{x^2+y^2}$ for $x$ is:
$\frac{x}{\sqrt{x^2+y^2}}+\frac{2 x-2}{\sqrt{x^2-2 x+y^2+1}}+\frac{2 x-6}{2 \sqrt{x^2-6 x+y^2-8 y+25}}$
with the following contour plot:

and its PDE for $y$ is given by:
$\frac{y}{\sqrt{x^2+y^2}}+\frac{2 y}{\sqrt{x^2-2 x+y^2+1}}+\frac{2 y-8}{2 \sqrt{x^2-6 x+y^2-8 y+25}}$
with the following contour plot:

And finally here is the intersection plot:

followed by the intersection plot where I removed the coefficient 2 in front of the square:

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