<s>[INST] Find the minimum of $\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$ Find the minimum of $$f(x,y)=\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$$ It seems that $f_x=f_y=0$ is very hard to compute. Is there any easier idea?
 [/INST] Since
$$f(x,y)= \sqrt{(2y-3)^2+1}+\dfrac{\sqrt2}2\sqrt{(6x-3)^2+1}+\dfrac12\sqrt{(6x-4y+2)^2+(6x)^2},$$
then the substitutions
$$2y-3 = \sinh s,\quad 6x-3=\sinh t,\quad s,t\in\mathbb R,\tag1$$
transform $\;f(x,y)\;$ to
$$g(s,t)= \cosh s + \dfrac{\sqrt2}{2} \cosh t+\dfrac12\,r(s,t),\tag2$$
where
$$r(s,t)=\sqrt{\left(\sinh t-2\sinh s-1\right)^2+\left(\sinh t+3\right)^2}.\tag3$$
The stationary points of $\;g(s,t)\;$ correspond to solutions of the system $\;g'_s=g'_t=0,\;$ or
\begin{cases}
\sinh s-\dfrac1{r(s,t)}\,\cosh s\,(\sinh t-2\sinh s-1)=0\\[4pt]
\dfrac{\sqrt2}2\sinh t-\dfrac1{r(s,t)}\,\cosh t\,(\sinh t-\sinh s+1)=0,\\[4pt]
\end{cases}
\begin{cases}
r(s,t) = \coth s\,(\sinh t-2\sinh s-1)\\[4pt]
r(s,t) = \sqrt2\,\coth t\,(\sinh t-\sinh s+1),\\[4pt]
\end{cases}
\begin{cases}
\big(\left(\sinh t-2\sinh s-1\right)^2+\left(\sinh t+3\right)^2\big)\sinh ^2s\\[4pt]
=(\sinh t-2\sinh s-1)^2(1+\sinh^2s)\\[4pt]
\big(\left(\sinh t-2\sinh s-1\right)^2+\left(\sinh t+3\right)^2\big)\sinh ^2t\\[4pt]
=2(\sinh t-\sinh s+1)^2(1+\sinh^2t)\\[4pt]
\coth s\,(\sinh t-2\sinh s-1)\ge0\\[4pt]
\coth t\,(\sinh t-\sinh s+1)\ge0,
\end{cases}
\begin{cases}
\left(\sinh t+3\right)\sinh s=\sinh t-2\sinh s-1\\[4pt]
\big(2\sinh^2t-4\sinh t \sinh s +4\sinh^2s+4\sinh t+4\sinh s+10\big)\sinh ^2t\\[4pt]
=2(\sinh^2t-2\sinh t \sinh s+\sinh^2s+2\sinh t-2\sinh s+1)(1+\sinh^2t)\\[4pt]
\sinh t+3\ge0,\\[4pt]
\coth t\,(\sinh t-\sinh s+1)\ge0,
\end{cases}
\begin{cases}
\sinh t+3 = \dfrac{2\sinh s+4}{1-\sinh s}\\[4pt]
(\sinh s+2)^2\sinh ^2t =(\sinh t-\sinh s+1)^2\\[4pt]
\sinh t\ge -3,\quad 1 > \sinh s \ge -2\\[4pt]
\sinh t\,(\sinh t-\sinh s+1)\ge0,
\end{cases}
\begin{cases}
\sinh t = \dfrac{5\sinh s+1}{1-\sinh s}\\[4pt]
(\sinh s+2)^2(5\sinh s+1)^2 =(5\sinh s+1+(\sinh s-1)^2)^2\\[4pt]
\sinh t\ge -3,\quad 1 > \sinh s \ge -2\\[4pt]
\sinh t\,(\sinh t-\sinh s+1)\ge0,
\end{cases}
\begin{cases}
\sinh t = \dfrac{5\sinh s+1}{1-\sinh s}\\[4pt]
(\sinh s+2)^2(5\sinh s+1)^2 =(\sin s+1)^2(\sinh s+2)^2\\[4pt]
\sinh t\ge -3,\quad 1 > \sinh s \ge -2\\[4pt]
\sinh t\,(\sinh t-\sinh s+1)\ge0,
\end{cases}
$$\begin{pmatrix}\sinh s\\ \sinh t\\ g(s,t)\end{pmatrix} =
\left\{\begin{pmatrix}-2\\ -3\\ 2\sqrt5\end{pmatrix},
\begin{pmatrix}-\frac13\\ -\frac12\\ \sqrt10 \end{pmatrix},
\begin{pmatrix}0 \\ 1\\ 4\end{pmatrix}\right\}.$$
Therefore, the least value of $\;f(x,y)\;$ is $\;\color{brown}{\mathbf{\sqrt{10}}}\;$ at $\;\color{brown}{\mathbf{(x,y)=\left(\dfrac5{12},\dfrac43\right)}}.\;$
</s>