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?

Complete the square in the first two terms and let $Y=2y-3$ and $X=3x-3/2$ to get $$f(X,Y)=\sqrt{Y^2+1}+\sqrt{2X^2+\frac12}+\sqrt{2\left(X+\frac12\right)^2+\left(Y+\frac12\right)^2-2XY+\frac74}$$ so that \begin{align}f_X&=\frac{2X}{\sqrt{2X^2+1/2}}+\frac{2X-Y+1}{\sqrt{2(X+1/2)^2+(Y+1/2)^2-2XY+7/4}}=0\\f_Y&=\frac Y{\sqrt{Y^2+1}}+\frac{Y-X+1/2}{\sqrt{2(X+1/2)^2+(Y+1/2)^2-2XY+7/4}}=0\end{align} which can be rewritten as \begin{align}\frac{4X^2}{2X^2+1/2}&=\frac{(2X-Y+1)^2}{2(X+1/2)^2+(Y+1/2)^2-2XY+7/4}\\\frac{Y^2}{Y^2+1}&=\frac{(Y-X+1/2)^2}{2(X+1/2)^2+(Y+1/2)^2-2XY+7/4}\end{align} and extensively simplified to give the symmetrical forms \begin{align}X^2(4Y^2+16Y+12)+4(Y-1)X-(Y-1)^2&=0\\Y^2(4X^2+12X+5)+4(2X-1)Y-(2X-1)^2&=0\end{align} on clearing denominators. Equating the two expressions on the left gives $$8X^2Y-6XY^2+8X^2-3Y^2-2XY-4X+3Y=0$$ which factorises to $(4X-3Y)(2XY+2X+Y-1)=0$, where the two interaction terms significantly aid this observation. Substituting $X=3Y/4$ gives the polynomial $$9Y^4+36Y^3+35Y^2-4Y-4=0$$ which has roots at $Y=-2,\pm1/3$ after using the rational root theorem. Hence $X=-3/2,\pm1/4$ and a minimum of $\sqrt{10}$ is obtained at $$(X,Y)=\left(-\frac14,-\frac13\right)\implies(x,y)=\left(\frac5{12},\frac43\right).$$ In fact, it is the global minimum as the expression $2XY+2X+Y-1=0$ is an anomaly due to squaring. The square root in the third term of $f(X,Y)$ means that the condition $f_X=0$ only holds if $X\le-3/2$ or $(X,Y)=(0,1)$ and $f_Y=0$ only holds if $-3/2\le X<0$ or $(X,Y)=(1/2,0)$.
