<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] Recall the well-known fact: If $g(u)$ is a convex function on $\mathbb{R}^n$, then
$$g(u) \ge g(v) + \nabla g(v)^\mathsf{T}(u - v), \ \forall u, v \in \mathbb{R}^n.$$
Clearly, $f(x, y)$ is a convex function on $\mathbb{R}^2$. Thus, we have
$$f(x, y) \ge f(\tfrac{5}{12}, \tfrac{4}{3}) + \tfrac{\partial f}{\partial x}(\tfrac{5}{12}, \tfrac{4}{3}) \cdot (x - \tfrac{5}{12})
+ \tfrac{\partial f}{\partial y}(\tfrac{5}{12}, \tfrac{4}{3}) \cdot (y - \tfrac{4}{3}) = \sqrt{10}.$$
Also, $f(\tfrac{5}{12}, \tfrac{4}{3}) = \sqrt{10}$.
The minimum of $f(x, y)$ is $\sqrt{10}$ achieved at $x = 5/12$ and $y = 4/3$.
Another way: Since $f(x, y)$ is a convex function on $\mathbb{R}^n$ and $\tfrac{\partial f}{\partial x}(\tfrac{5}{12}, \tfrac{4}{3})
= \tfrac{\partial f}{\partial y}(\tfrac{5}{12}, \tfrac{4}{3}) = 0$, $f(x,y)$ achieves its minimum at $x = 5/12$ and $y = 4/3$.
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