Find the minimum of sum Let $A(1, 1)$ and $B(3, 2)$ be two points. Find point $M(x, 0)$ so that the sum $AM + MB$ is minimum. After that, find the minimum of the sum.
In order for the sum to be minimum, I concluded that $x$ must lay between $1$ and $3$. I also calculated $AM$ and $MB$:
$$AM = \sqrt{x^2 - 2x + 2}$$
$$BM = \sqrt{x^2 - 6x + 13}$$
I don't now how to proceed from this point, so I would appreciate any help from you!
Thank you in advance!
 A: Consider $B'(3,-2)$. Let $N$ be the intersection point of $x$-axis with $AB'$.
Then,
$$|AM|+|BM|=|AM|+|B'M|\ge |AN|+|B'N|$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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Let's $\ds{\quad x - 1 = \tan\pars{\alpha}\quad}$ and $\ds{\quad x - 3 = 2\tan\pars{\beta}\quad}$ such that
\begin{align}
\,\mathrm{D}\pars{x} & \equiv
\root{\pars{x - 1}^{2} + 1} + \root{\pars{x - 3}^{2} + 4} =
\sec\pars{\alpha} + 2\sec\pars{\beta}
\end{align}
$\ds{\alpha}$ and $\ds{\beta}$ are constrained by $\ds{2 = \pars{x - 1} - \pars{x - 3} = \tan\pars{\alpha} - 2\tan\pars{\beta}}$. So, we have to minimises
$\ds{\sec\pars{\alpha} + 2\sec\pars{\beta}}$ with the above constraining relation. Namely,
$$
\sec\pars{\alpha} + 2\sec\pars{\beta} -
\mu\bracks{\tan\pars{\alpha} - 2\tan\pars{\beta} - 2}
$$
$\ds{\mu}$ is a Lagrange Multiplier.
$$
\left.\begin{array}{rcrcl}
\ds{\sec\pars{\alpha}\tan\pars{\alpha}} & \ds{-} & \ds{\mu\sec^{2}\pars{\alpha}}
& \ds{=} & \ds{0}
\\[1mm]
\ds{2\sec\pars{\beta}\tan\pars{\beta}} & \ds{+} & \ds{2\mu\sec^{2}\pars{\beta}}
& \ds{=} & \ds{0}
\end{array}\right\rbrace
\quad\imp\quad
\left\lbrace\begin{array}{rcr}
\ds{\sin\pars{\alpha}} & \ds{=} & \ds{\mu}
\\[1mm]
\ds{\sin\pars{\beta}} & \ds{=} & \ds{-\mu}
\end{array}\right.
$$

Then, $\ds{\tan\pars{\alpha} = \pm\tan\pars{\beta}}$:


*

*$\ds{\tan\pars{\alpha} = \tan\pars{\beta}\ \imp\ x - 1 = {x - 3 \over 2}\
\imp\ x = -1\ \imp\ \,\mathrm{D}\pars{-1} = 3\root{5} \approx 6.7082}$.

*$\ds{\tan\pars{\alpha} = -\tan\pars{\beta}\ \imp\ x - 1 =
-\,{x - 3 \over 2}\ \imp\ x = {5 \over 3}\ \imp\ \,\mathrm{D}\pars{5 \over 3} = \root{13} \approx 3.6056}$.

$$
\color{#f00}{Solution:\quad\,\mathrm{M} = \pars{{5 \over 3},0}}\quad
\mbox{with distance sum to }\ \,\mathrm{A}\ \mbox{and}\ \,\mathrm{B}\
\mbox{equal to}\ \color{#f00}{\root{13}} \approx 3.6056.
$$

